• 6.2 Graphs of the Other Trigonometric Functions
  • Introduction to Functions
  • 1.1 Functions and Function Notation
  • 1.2 Domain and Range
  • 1.3 Rates of Change and Behavior of Graphs
  • 1.4 Composition of Functions
  • 1.5 Transformation of Functions
  • 1.6 Absolute Value Functions
  • 1.7 Inverse Functions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • Practice Test
  • Introduction to Linear Functions
  • 2.1 Linear Functions
  • 2.2 Graphs of Linear Functions
  • 2.3 Modeling with Linear Functions
  • 2.4 Fitting Linear Models to Data
  • Introduction to Polynomial and Rational Functions
  • 3.1 Complex Numbers
  • 3.2 Quadratic Functions
  • 3.3 Power Functions and Polynomial Functions
  • 3.4 Graphs of Polynomial Functions
  • 3.5 Dividing Polynomials
  • 3.6 Zeros of Polynomial Functions
  • 3.7 Rational Functions
  • 3.8 Inverses and Radical Functions
  • 3.9 Modeling Using Variation
  • Introduction to Exponential and Logarithmic Functions
  • 4.1 Exponential Functions
  • 4.2 Graphs of Exponential Functions
  • 4.3 Logarithmic Functions
  • 4.4 Graphs of Logarithmic Functions
  • 4.5 Logarithmic Properties
  • 4.6 Exponential and Logarithmic Equations
  • 4.7 Exponential and Logarithmic Models
  • 4.8 Fitting Exponential Models to Data
  • Introduction to Trigonometric Functions
  • 5.2 Unit Circle: Sine and Cosine Functions
  • 5.3 The Other Trigonometric Functions
  • 5.4 Right Triangle Trigonometry
  • Introduction to Periodic Functions
  • 6.1 Graphs of the Sine and Cosine Functions
  • 6.3 Inverse Trigonometric Functions
  • Introduction to Trigonometric Identities and Equations
  • 7.1 Solving Trigonometric Equations with Identities
  • 7.2 Sum and Difference Identities
  • 7.3 Double-Angle, Half-Angle, and Reduction Formulas
  • 7.4 Sum-to-Product and Product-to-Sum Formulas
  • 7.5 Solving Trigonometric Equations
  • 7.6 Modeling with Trigonometric Functions
  • Introduction to Further Applications of Trigonometry
  • 8.1 Non-right Triangles: Law of Sines
  • 8.2 Non-right Triangles: Law of Cosines
  • 8.3 Polar Coordinates
  • 8.4 Polar Coordinates: Graphs
  • 8.5 Polar Form of Complex Numbers
  • 8.6 Parametric Equations
  • 8.7 Parametric Equations: Graphs
  • 8.8 Vectors
  • Introduction to Systems of Equations and Inequalities
  • 9.1 Systems of Linear Equations: Two Variables
  • 9.2 Systems of Linear Equations: Three Variables
  • 9.3 Systems of Nonlinear Equations and Inequalities: Two Variables
  • 9.4 Partial Fractions
  • 9.5 Matrices and Matrix Operations
  • 9.6 Solving Systems with Gaussian Elimination
  • 9.7 Solving Systems with Inverses
  • 9.8 Solving Systems with Cramer's Rule
  • Introduction to Analytic Geometry
  • 10.1 The Ellipse
  • 10.2 The Hyperbola
  • 10.3 The Parabola
  • 10.4 Rotation of Axes
  • 10.5 Conic Sections in Polar Coordinates
  • Introduction to Sequences, Probability and Counting Theory
  • 11.1 Sequences and Their Notations
  • 11.2 Arithmetic Sequences
  • 11.3 Geometric Sequences
  • 11.4 Series and Their Notations
  • 11.5 Counting Principles
  • 11.6 Binomial Theorem
  • 11.7 Probability
  • Introduction to Calculus
  • 12.1 Finding Limits: Numerical and Graphical Approaches
  • 12.2 Finding Limits: Properties of Limits
  • 12.3 Continuity
  • 12.4 Derivatives
  • A | Basic Functions and Identities

Learning Objectives

In this section, you will:

  • Analyze the graph of  y=tan x.
  • Graph variations of  y=tan x.
  • Analyze the graphs of  y=sec x  and  y=csc x.
  • Graph variations of  y=sec x  and  y=csc x.
  • Analyze the graph of  y=cot x.
  • Graph variations of  y=cot x.

We know the tangent function can be used to find distances, such as the height of a building, mountain, or flagpole. But what if we want to measure repeated occurrences of distance? Imagine, for example, a fire truck parked next to a warehouse. The rotating light from the truck would travel across the wall of the warehouse in regular intervals. If the input is time, the output would be the distance the beam of light travels. The beam of light would repeat the distance at regular intervals. The tangent function can be used to approximate this distance. Asymptotes would be needed to illustrate the repeated cycles when the beam runs parallel to the wall because, seemingly, the beam of light could appear to extend forever. The graph of the tangent function would clearly illustrate the repeated intervals. In this section, we will explore the graphs of the tangent and other trigonometric functions.

Analyzing the Graph of y = tan x

We will begin with the graph of the tangent function, plotting points as we did for the sine and cosine functions. Recall that

The period of the tangent function is π π because the graph repeats itself on intervals of k π k π where k k is a constant. If we graph the tangent function on − π 2 − π 2 to π 2 , π 2 , we can see the behavior of the graph on one complete cycle. If we look at any larger interval, we will see that the characteristics of the graph repeat.

We can determine whether tangent is an odd or even function by using the definition of tangent.

Therefore, tangent is an odd function. We can further analyze the graphical behavior of the tangent function by looking at values for some of the special angles, as listed in Table 1 .

These points will help us draw our graph, but we need to determine how the graph behaves where it is undefined. If we look more closely at values when π 3 < x < π 2 , π 3 < x < π 2 , we can use a table to look for a trend. Because π 3 ≈ 1.05 π 3 ≈ 1.05 and π 2 ≈ 1.57 , π 2 ≈ 1.57 , we will evaluate x x at radian measures 1.05 < x < 1.57 1.05 < x < 1.57 as shown in Table 2 .

As x x approaches π 2 , π 2 , the outputs of the function get larger and larger. Because y = tan x y = tan x is an odd function, we see the corresponding table of negative values in Table 3 .

We can see that, as x x approaches − π 2 , − π 2 , the outputs get smaller and smaller. Remember that there are some values of x x for which cos x = 0. cos x = 0. For example, cos ( π 2 ) = 0 cos ( π 2 ) = 0 and cos ( 3 π 2 ) = 0. cos ( 3 π 2 ) = 0. At these values, the tangent function is undefined, so the graph of y = tan x y = tan x has discontinuities at x = π 2  and  3 π 2 . x = π 2  and  3 π 2 . At these values, the graph of the tangent has vertical asymptotes. Figure 1 represents the graph of y = tan x . y = tan x . The tangent is positive from 0 to π 2 π 2 and from π π to 3 π 2 , 3 π 2 , corresponding to quadrants I and III of the unit circle.

Graphing Variations of y = tan x

As with the sine and cosine functions, the tangent function can be described by a general equation.

We can identify horizontal and vertical stretches and compressions using values of A A and B . B . The horizontal stretch can typically be determined from the period of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph.

Because there are no maximum or minimum values of a tangent function, the term amplitude cannot be interpreted as it is for the sine and cosine functions. Instead, we will use the phrase stretching/compressing factor when referring to the constant A . A .

Features of the Graph of y = A tan( Bx )

  • The stretching factor is | A | . | A | .
  • The period is P = π | B | . P = π | B | .
  • The domain is all real numbers x , x , where x ≠ π 2 | B | + π | B | k x ≠ π 2 | B | + π | B | k such that k k is an integer.
  • The range is ( −∞ , ∞ ) . ( −∞ , ∞ ) .
  • The asymptotes occur at x = π 2 | B | + π | B | k , x = π 2 | B | + π | B | k , where k k is an integer.
  • y = A tan ( B x ) y = A tan ( B x ) is an odd function.

Graphing One Period of a Stretched or Compressed Tangent Function

We can use what we know about the properties of the tangent function to quickly sketch a graph of any stretched and/or compressed tangent function of the form f ( x ) = A tan ( B x ) . f ( x ) = A tan ( B x ) . We focus on a single period of the function including the origin, because the periodic property enables us to extend the graph to the rest of the function’s domain if we wish. Our limited domain is then the interval ( − P 2 , P 2 ) ( − P 2 , P 2 ) and the graph has vertical asymptotes at ± P 2 ± P 2 where P = π B . P = π B . On ( − π 2 , π 2 ) , ( − π 2 , π 2 ) , the graph will come up from the left asymptote at x = − π 2 , x = − π 2 , cross through the origin, and continue to increase as it approaches the right asymptote at x = π 2 . x = π 2 . To make the function approach the asymptotes at the correct rate, we also need to set the vertical scale by actually evaluating the function for at least one point that the graph will pass through. For example, we can use

because tan ( π 4 ) = 1. tan ( π 4 ) = 1.

Given the function f ( x ) = A tan ( B x ) , f ( x ) = A tan ( B x ) , graph one period.

  • Identify the stretching factor, | A | . | A | .
  • Identify B B and determine the period, P = π | B | . P = π | B | .
  • Draw vertical asymptotes at x = − P 2 x = − P 2 and x = P 2 . x = P 2 .
  • For A B > 0 , A B > 0 , the graph approaches the left asymptote at negative output values and the right asymptote at positive output values (reverse for A B < 0 A B < 0 ).
  • Plot reference points at ( P 4 , A ) , ( P 4 , A ) , ( 0 , 0 ) , ( 0 , 0 ) , and ( − P 4 ,− A ) , ( − P 4 ,− A ) , and draw the graph through these points.

Sketching a Compressed Tangent

Sketch a graph of one period of the function y = 0.5 tan ( π 2 x ) . y = 0.5 tan ( π 2 x ) .

First, we identify A A and B . B .

Because A = 0.5 A = 0.5 and B = π 2 , B = π 2 , we can find the stretching/compressing factor and period. The period is π π 2 = 2 , π π 2 = 2 , so the asymptotes are at x = ± 1. x = ± 1. At a quarter period from the origin, we have

This means the curve must pass through the points ( 0.5 , 0.5 ) , ( 0.5 , 0.5 ) , ( 0 , 0 ) , ( 0 , 0 ) , and ( − 0.5 , −0.5 ) . ( − 0.5 , −0.5 ) . The only inflection point is at the origin. Figure 2 shows the graph of one period of the function.

Sketch a graph of f ( x ) = 3 tan ( π 6 x ) . f ( x ) = 3 tan ( π 6 x ) .

Graphing One Period of a Shifted Tangent Function

Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we add C C and D D to the general form of the tangent function.

The graph of a transformed tangent function is different from the basic tangent function tan x tan x in several ways:

Features of the Graph of y = A tan( Bx − C )+ D

  • The period is π | B | . π | B | .
  • The domain is x ≠ C B + π | B | k , x ≠ C B + π | B | k , where k k is an integer.
  • The vertical asymptotes occur at x = C B + π 2 | B | k , x = C B + π 2 | B | k , where k k is an odd integer.
  • There is no amplitude.

Given the function y = A tan ( B x − C ) + D , y = A tan ( B x − C ) + D , sketch the graph of one period.

  • Express the function given in the form y = A tan ( B x − C ) + D . y = A tan ( B x − C ) + D .
  • Identify the stretching/compressing factor , | A | . | A | .
  • Identify C C and determine the phase shift, C B . C B .
  • Draw the graph of y = A tan ( B x ) y = A tan ( B x ) shifted to the right by C B C B and up by D . D .
  • Sketch the vertical asymptotes, which occur at x = C B + π 2 | B | k , x = C B + π 2 | B | k , where k k is an odd integer.
  • Plot any three reference points and draw the graph through these points.

Graph one period of the function y = −2 tan ( π x + π ) −1. y = −2 tan ( π x + π ) −1.

  • Step 1. The function is already written in the form y = A tan ( B x − C ) + D . y = A tan ( B x − C ) + D .
  • Step 2. A = −2 , A = −2 , so the stretching factor is | A | = 2. | A | = 2.
  • Step 3. B = π , B = π , so the period is P = π | B | = π π = 1. P = π | B | = π π = 1.
  • Step 4. C = − π , C = − π , so the phase shift is C B = − π π = −1. C B = − π π = −1.

Note that this is a decreasing function because A < 0. A < 0.

How would the graph in Example 2 look different if we made A = 2 A = 2 instead of −2 ? −2 ?

Given the graph of a tangent function, identify horizontal and vertical stretches.

  • Find the period P P from the spacing between successive vertical asymptotes or x -intercepts.
  • Write f ( x ) = A tan ( π P x ) . f ( x ) = A tan ( π P x ) .
  • Determine a convenient point ( x , f ( x ) ) ( x , f ( x ) ) on the given graph and use it to determine A . A .

Identifying the Graph of a Stretched Tangent

Find a formula for the function graphed in Figure 4 .

The graph has the shape of a tangent function.

  • Step 1. One cycle extends from –4 to 4, so the period is P = 8. P = 8. Since P = π | B | , P = π | B | , we have B = π P = π 8 . B = π P = π 8 .
  • Step 2. The equation must have the form f ( x ) = A tan ( π 8 x ) . f ( x ) = A tan ( π 8 x ) .
  • Step 3. To find the vertical stretch A , A , we can use the point ( 2 , 2 ) . ( 2 , 2 ) . 2 = A tan ( π 8 ⋅ 2 ) = A tan ( π 4 ) 2 = A tan ( π 8 ⋅ 2 ) = A tan ( π 4 )

Because tan ( π 4 ) = 1 , tan ( π 4 ) = 1 , A = 2. A = 2.

This function would have a formula f ( x ) = 2 tan ( π 8 x ) . f ( x ) = 2 tan ( π 8 x ) .

Find a formula for the function in Figure 5 .

Analyzing the Graphs of y = sec x and y = csc x

The secant was defined by the reciprocal identity sec x = 1 cos x . sec x = 1 cos x . Notice that the function is undefined when the cosine is 0, leading to vertical asymptotes at π 2 , π 2 , 3 π 2 , 3 π 2 , etc. Because the cosine is never more than 1 in absolute value, the secant, being the reciprocal, will never be less than 1 in absolute value.

We can graph y = sec x y = sec x by observing the graph of the cosine function because these two functions are reciprocals of one another. See Figure 6 . The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined.

The secant graph has vertical asymptotes at each value of x x where the cosine graph crosses the x -axis; we show these in the graph below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant.

Note that, because cosine is an even function, secant is also an even function. That is, sec ( − x ) = sec x . sec ( − x ) = sec x .

As we did for the tangent function, we will again refer to the constant | A | | A | as the stretching factor, not the amplitude.

Features of the Graph of y = A sec( Bx )

  • The period is 2 π | B | . 2 π | B | .
  • The domain is x ≠ π 2 | B | k , x ≠ π 2 | B | k , where k k is an odd integer.
  • The range is ( − ∞ , − | A | ] ∪ [ | A | , ∞ ) . ( − ∞ , − | A | ] ∪ [ | A | , ∞ ) .
  • The vertical asymptotes occur at x = π 2 | B | k , x = π 2 | B | k , where k k is an odd integer.
  • y = A sec ( B x ) y = A sec ( B x ) is an even function because cosine is an even function.

Similar to the secant, the cosecant is defined by the reciprocal identity csc x = 1 sin x . csc x = 1 sin x . Notice that the function is undefined when the sine is 0, leading to a vertical asymptote in the graph at 0 , 0 , π , π , etc. Since the sine is never more than 1 in absolute value, the cosecant, being the reciprocal, will never be less than 1 in absolute value.

We can graph y = csc x y = csc x by observing the graph of the sine function because these two functions are reciprocals of one another. See Figure 7 . The graph of sine is shown as a dashed orange wave so we can see the relationship. Where the graph of the sine function decreases, the graph of the cosecant function increases. Where the graph of the sine function increases, the graph of the cosecant function decreases.

The cosecant graph has vertical asymptotes at each value of x x where the sine graph crosses the x -axis; we show these in the graph below with dashed vertical lines.

Note that, since sine is an odd function, the cosecant function is also an odd function. That is, csc ( − x ) = −csc x . csc ( − x ) = −csc x .

The graph of cosecant, which is shown in Figure 7 , is similar to the graph of secant.

Features of the Graph of y = A csc( Bx )

  • The domain is x ≠ π | B | k , x ≠ π | B | k , where k k is an integer.
  • The asymptotes occur at x = π | B | k , x = π | B | k , where k k is an integer.
  • y = A csc ( B x ) y = A csc ( B x ) is an odd function because sine is an odd function.

Graphing Variations of y = sec x and y = csc x

For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function in the same way as for the secant and other functions.The equations become the following.

Features of the Graph of y = A sec( Bx − C )+ D

  • The domain is x ≠ C B + π 2 | B | k , x ≠ C B + π 2 | B | k , where k k is an odd integer.
  • The range is ( − ∞ , − | A | + D ] ∪ [ | A | + D , ∞ ) . ( − ∞ , − | A | + D ] ∪ [ | A | + D , ∞ ) .
  • y = A sec ( B x - C ) + D y = A sec ( B x - C ) + D is an even function because cosine is an even function.

Features of the Graph of y = A csc( Bx − C )+ D

  • The vertical asymptotes occur at x = C B + π | B | k , x = C B + π | B | k , where k k is an integer.
  • y = A csc ( B x - C ) + D y = A csc ( B x - C ) + D is an odd function because sine is an odd function.

Given a function of the form y = A sec ( B x ) , y = A sec ( B x ) , graph one period.

  • Express the function given in the form y = A sec ( B x ) . y = A sec ( B x ) .
  • Identify the stretching/compressing factor, | A | . | A | .
  • Identify B B and determine the period, P = 2 π | B | . P = 2 π | B | .
  • Sketch the graph of y = A cos ( B x ) . y = A cos ( B x ) .
  • Use the reciprocal relationship between y = cos x y = cos x and y = sec x y = sec x to draw the graph of y = A sec ( B x ) . y = A sec ( B x ) .
  • Sketch the asymptotes.
  • Plot any two reference points and draw the graph through these points.

Graphing a Variation of the Secant Function

Graph one period of f ( x ) = 2.5 sec ( 0.4 x ) . f ( x ) = 2.5 sec ( 0.4 x ) .

  • Step 1. The given function is already written in the general form, y = A sec ( B x ) . y = A sec ( B x ) .
  • Step 2. A = 2.5 A = 2.5 so the stretching factor is 2 .5 . 2 .5 .
  • Step 3. B = 0.4 B = 0.4 so P = 2 π 0.4 = 5 π . P = 2 π 0.4 = 5 π . The period is 5 π 5 π units.
  • Step 4. Sketch the graph of the function g ( x ) = 2.5 cos ( 0.4 x ) . g ( x ) = 2.5 cos ( 0.4 x ) .
  • Step 5. Use the reciprocal relationship of the cosine and secant functions to draw the cosecant function.

Graph one period of f ( x ) = − 2.5 sec ( 0.4 x ) . f ( x ) = − 2.5 sec ( 0.4 x ) .

Do the vertical shift and stretch/compression affect the secant’s range?

Yes. The range of f ( x ) = A sec ( B x − C ) + D f ( x ) = A sec ( B x − C ) + D is ( − ∞ , − | A | + D ] ∪ [ | A | + D , ∞ ) . ( − ∞ , − | A | + D ] ∪ [ | A | + D , ∞ ) .

Given a function of the form f ( x ) = A sec ( B x − C ) + D , f ( x ) = A sec ( B x − C ) + D , graph one period.

  • Express the function given in the form y = A sec ( B x − C ) + D . y = A sec ( B x − C ) + D .
  • Identify B B and determine the period, 2 π | B | . 2 π | B | .
  • Draw the graph of y = A sec ( B x ) y = A sec ( B x ) , but shift it to the right by C B C B and up by D . D .

Graph one period of y = 4 sec ( π 3 x − π 2 ) + 1. y = 4 sec ( π 3 x − π 2 ) + 1.

  • Step 1. Express the function given in the form y = 4 sec ( π 3 x − π 2 ) + 1. y = 4 sec ( π 3 x − π 2 ) + 1.
  • Step 2. The stretching/compressing factor is | A | = 4. | A | = 4.
  • Step 3. The period is 2 π | B | = 2 π π 3        = 2 π 1 ⋅ 3 π        = 6 2 π | B | = 2 π π 3        = 2 π 1 ⋅ 3 π        = 6
  • Step 4. The phase shift is C B = π 2 π 3     = π 2 ⋅ 3 π     = 1.5 C B = π 2 π 3     = π 2 ⋅ 3 π     = 1.5
  • Step 5. Draw the graph of y = A sec ( B x ) , y = A sec ( B x ) , but shift it to the right by C B = 1.5 C B = 1.5 and up by D = 1. D = 1.
  • Step 6. Sketch the vertical asymptotes, which occur at x = 0 , x = 3 , x = 0 , x = 3 , and x = 6. x = 6. There is a local minimum at ( 1.5 , 5 ) ( 1.5 , 5 ) and a local maximum at ( 4.5 , − 3 ) . ( 4.5 , − 3 ) . Figure 9 shows the graph.

Graph one period of f ( x ) = − 6 sec ( 4 x + 2 ) − 8. f ( x ) = − 6 sec ( 4 x + 2 ) − 8.

The domain of csc x csc x was given to be all x x such that x ≠ k π x ≠ k π for any integer k . k . Would the domain of y = A csc ( B x − C ) + D be x ≠ C + k π B ? y = A csc ( B x − C ) + D be x ≠ C + k π B ?

Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original function’s input.

Given a function of the form y = A csc ( B x ) , y = A csc ( B x ) , graph one period.

  • Express the function given in the form y = A csc ( B x ) . y = A csc ( B x ) .
  • | A | . | A | .
  • Draw the graph of y = A sin ( B x ) . y = A sin ( B x ) .
  • Use the reciprocal relationship between y = sin x y = sin x and y = csc x y = csc x to draw the graph of y = A csc ( B x ) . y = A csc ( B x ) .

Graphing a Variation of the Cosecant Function

Graph one period of f ( x ) = −3 csc ( 4 x ) . f ( x ) = −3 csc ( 4 x ) .

  • Step 1. The given function is already written in the general form, y = A csc ( B x ) . y = A csc ( B x ) .
  • Step 2. | A | = | − 3 | = 3 , | A | = | − 3 | = 3 , so the stretching factor is 3.
  • Step 3. B = 4 , B = 4 , so P = 2 π 4 = π 2 . P = 2 π 4 = π 2 . The period is π 2 π 2 units.
  • Step 4. Sketch the graph of the function g ( x ) = −3 sin ( 4 x ) . g ( x ) = −3 sin ( 4 x ) .
  • Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function .

Graph one period of f ( x ) = 0.5 csc ( 2 x ) . f ( x ) = 0.5 csc ( 2 x ) .

Given a function of the form f ( x ) = A csc ( B x − C ) + D , f ( x ) = A csc ( B x − C ) + D , graph one period.

  • Express the function given in the form y = A csc ( B x − C ) + D . y = A csc ( B x − C ) + D .
  • Draw the graph of y = A csc ( B x ) y = A csc ( B x ) but shift it to the right by C B C B and up by D . D .
  • Sketch the vertical asymptotes, which occur at x = C B + π | B | k , x = C B + π | B | k , where k k is an integer.

Graphing a Vertically Stretched, Horizontally Compressed, and Vertically Shifted Cosecant

Sketch a graph of y = 2 csc ( π 2 x ) + 1. y = 2 csc ( π 2 x ) + 1. What are the domain and range of this function?

  • Step 1. Express the function given in the form y = 2 csc ( π 2 x ) + 1. y = 2 csc ( π 2 x ) + 1.
  • Step 2. Identify the stretching/compressing factor, | A | = 2. | A | = 2.
  • Step 3. The period is 2 π | B | = 2 π π 2 = 2 π 1 ⋅ 2 π = 4. 2 π | B | = 2 π π 2 = 2 π 1 ⋅ 2 π = 4.
  • Step 4. The phase shift is 0 π 2 = 0. 0 π 2 = 0.
  • Step 5. Draw the graph of y = A csc ( B x ) y = A csc ( B x ) but shift it up D = 1. D = 1.
  • Step 6. Sketch the vertical asymptotes, which occur at x = 0 , x = 2 , x = 4. x = 0 , x = 2 , x = 4.

The graph for this function is shown in Figure 11 .

The vertical asymptotes shown on the graph mark off one period of the function, and the local extrema in this interval are shown by dots. Notice how the graph of the transformed cosecant relates to the graph of f ( x ) = 2 sin ( π 2 x ) + 1 , f ( x ) = 2 sin ( π 2 x ) + 1 , shown as the orange dashed wave.

Given the graph of f ( x ) = 2 cos ( π 2 x ) + 1 f ( x ) = 2 cos ( π 2 x ) + 1 shown in Figure 12 , sketch the graph of g ( x ) = 2 sec ( π 2 x ) + 1 g ( x ) = 2 sec ( π 2 x ) + 1 on the same axes.

Analyzing the Graph of y = cot x

The last trigonometric function we need to explore is cotangent . The cotangent is defined by the reciprocal identity cot x = 1 tan x . cot x = 1 tan x . Notice that the function is undefined when the tangent function is 0, leading to a vertical asymptote in the graph at 0 , π , 0 , π , etc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers.

We can graph y = cot x y = cot x by observing the graph of the tangent function because these two functions are reciprocals of one another. See Figure 13 . Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of the tangent function increases, the graph of the cotangent function decreases.

The cotangent graph has vertical asymptotes at each value of x x where tan x = 0 ; tan x = 0 ; we show these in the graph below with dashed lines. Since the cotangent is the reciprocal of the tangent, cot x cot x has vertical asymptotes at all values of x x where tan x = 0 , tan x = 0 , and cot x = 0 cot x = 0 at all values of x x where tan x tan x has its vertical asymptotes.

Features of the Graph of y = A cot( Bx )

  • The range is ( − ∞ , ∞ ) . ( − ∞ , ∞ ) .
  • y = A cot ( B x ) y = A cot ( B x ) is an odd function.

Graphing Variations of y = cot x

We can transform the graph of the cotangent in much the same way as we did for the tangent. The equation becomes the following.

Features of the Graph of y = A cot( Bx −C)+ D

  • y = A cot ( B x ) y = A cot ( B x ) is an odd function because it is the quotient of even and odd functions (cosine and sine, respectively)

Given a modified cotangent function of the form f ( x ) = A cot ( B x ) , f ( x ) = A cot ( B x ) , graph one period.

  • Express the function in the form f ( x ) = A cot ( B x ) . f ( x ) = A cot ( B x ) .
  • Identify the period, P = π | B | . P = π | B | .
  • Draw the graph of y = A tan ( B x ) . y = A tan ( B x ) .
  • Plot any two reference points.
  • Use the reciprocal relationship between tangent and cotangent to draw the graph of y = A cot ( B x ) . y = A cot ( B x ) .

Graphing Variations of the Cotangent Function

Determine the stretching factor, period, and phase shift of y = 3 cot ( 4 x ) , y = 3 cot ( 4 x ) , and then sketch a graph.

  • Step 1. Expressing the function in the form f ( x ) = A cot ( B x ) f ( x ) = A cot ( B x ) gives f ( x ) = 3 cot ( 4 x ) . f ( x ) = 3 cot ( 4 x ) .
  • Step 2. The stretching factor is | A | = 3. | A | = 3.
  • Step 3. The period is P = π 4 . P = π 4 .
  • Step 4. Sketch the graph of y = 3 tan ( 4 x ) . y = 3 tan ( 4 x ) .
  • Step 5. Plot two reference points. Two such points are ( π 16 , 3 ) ( π 16 , 3 ) and ( 3 π 16 , −3 ) . ( 3 π 16 , −3 ) .
  • Step 6. Use the reciprocal relationship to draw y = 3 cot ( 4 x ) . y = 3 cot ( 4 x ) .
  • Step 7. Sketch the asymptotes, x = 0 , x = π 4 . x = 0 , x = π 4 .

The blue graph in Figure 14 shows y = 3 tan ( 4 x ) y = 3 tan ( 4 x ) and the green graph shows y = 3 cot ( 4 x ) . y = 3 cot ( 4 x ) .

Given a modified cotangent function of the form f ( x ) = A cot ( B x − C ) + D , f ( x ) = A cot ( B x − C ) + D , graph one period.

  • Express the function in the form f ( x ) = A cot ( B x − C ) + D . f ( x ) = A cot ( B x − C ) + D .
  • Identify the phase shift, C B . C B .
  • Sketch the asymptotes x = C B + π | B | k , x = C B + π | B | k , where k k is an integer.

Graphing a Modified Cotangent

Sketch a graph of one period of the function f ( x ) = 4 cot ( π 8 x − π 2 ) − 2. f ( x ) = 4 cot ( π 8 x − π 2 ) − 2.

  • Step 1. The function is already written in the general form f ( x ) = A cot ( B x − C ) + D . f ( x ) = A cot ( B x − C ) + D .
  • Step 2. A = 4 , A = 4 , so the stretching factor is 4.
  • Step 3. B = π 8 , B = π 8 , so the period is P = π | B | = π π 8 = 8. P = π | B | = π π 8 = 8.
  • Step 4. C = π 2 , C = π 2 , so the phase shift is C B = π 2 π 8 = 4. C B = π 2 π 8 = 4.
  • Step 5. We draw f ( x ) = 4 tan ( π 8 x − π 2 ) − 2. f ( x ) = 4 tan ( π 8 x − π 2 ) − 2.
  • Step 6-7. Three points we can use to guide the graph are ( 6 , 2 ) , ( 8 , − 2 ) , ( 6 , 2 ) , ( 8 , − 2 ) , and ( 10 , − 6 ) . ( 10 , − 6 ) . We use the reciprocal relationship of tangent and cotangent to draw f ( x ) = 4 cot ( π 8 x − π 2 ) − 2. f ( x ) = 4 cot ( π 8 x − π 2 ) − 2.
  • Step 8. The vertical asymptotes are x = 4 x = 4 and x = 12. x = 12.

The graph is shown in Figure 15 .

Using the Graphs of Trigonometric Functions to Solve Real-World Problems

Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a fire truck and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function .

Using Trigonometric Functions to Solve Real-World Scenarios

Suppose the function y = 5 tan ( π 4 t ) y = 5 tan ( π 4 t ) marks the distance in the movement of a light beam from the top of a police car across a wall where t t is the time in seconds and y y is the distance in feet from a point on the wall directly across from the police car.

  • ⓐ Find and interpret the stretching factor and period.
  • ⓑ Graph on the interval [ 0 , 5 ] . [ 0 , 5 ] .
  • ⓒ Evaluate f ( 1 ) f ( 1 ) and discuss the function’s value at that input.

We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period.

The period is π π 4 = π 1 ⋅ 4 π = 4. π π 4 = π 1 ⋅ 4 π = 4. This means that every 4 seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches.

  • ⓒ period: f ( 1 ) = 5 tan ( π 4 ( 1 ) ) = 5 ( 1 ) = 5 ; f ( 1 ) = 5 tan ( π 4 ( 1 ) ) = 5 ( 1 ) = 5 ; after 1 second, the beam of has moved 5 ft from the spot across from the police car.

Access these online resources for additional instruction and practice with graphs of other trigonometric functions.

  • Graphing the Tangent
  • Graphing Cosecant and Secant
  • Graphing the Cotangent

Explain how the graph of the sine function can be used to graph y = csc x . y = csc x .

How can the graph of y = cos x y = cos x be used to construct the graph of y = sec x ? y = sec x ?

Explain why the period of tan x tan x is equal to π . π .

Why are there no intercepts on the graph of y = csc x ? y = csc x ?

How does the period of y = csc x y = csc x compare with the period of y = sin x ? y = sin x ?

For the following exercises, match each trigonometric function with one of the following graphs.

f ( x ) = tan x f ( x ) = tan x

f ( x ) = sec x f ( x ) = sec x

f ( x ) = csc x f ( x ) = csc x

f ( x ) = cot x f ( x ) = cot x

For the following exercises, find the period and horizontal shift of each of the functions.

f ( x ) = 2 tan ( 4 x − 32 ) f ( x ) = 2 tan ( 4 x − 32 )

h ( x ) = 2 sec ( π 4 ( x + 1 ) ) h ( x ) = 2 sec ( π 4 ( x + 1 ) )

m ( x ) = 6 csc ( π 3 x + π ) m ( x ) = 6 csc ( π 3 x + π )

For the following exercises, evaluate the transformed functions.

If tan x = − 1.5 , tan x = − 1.5 , find tan ( − x ) . tan ( − x ) .

If sec x = 2 , sec x = 2 , find sec ( − x ) . sec ( − x ) .

If csc x = − 5 , csc x = − 5 , find csc ( − x ) . csc ( − x ) .

If x sin x = 2 , x sin x = 2 , find ( − x ) sin ( − x ) . ( − x ) sin ( − x ) .

For the following exercises, rewrite each expression such that the argument x x is positive.

cot ( − x ) cos ( − x ) + sin ( − x ) cot ( − x ) cos ( − x ) + sin ( − x )

cos ( − x ) + tan ( − x ) sin ( − x ) cos ( − x ) + tan ( − x ) sin ( − x )

For the following exercises, sketch two periods of the graph for each of the following functions. Identify the stretching factor, period, and asymptotes.

j ( x ) = tan ( π 2 x ) j ( x ) = tan ( π 2 x )

p ( x ) = tan ( x − π 2 ) p ( x ) = tan ( x − π 2 )

f ( x ) = 4 tan ( x ) f ( x ) = 4 tan ( x )

f ( x ) = tan ( x + π 4 ) f ( x ) = tan ( x + π 4 )

f ( x ) = π tan ( π x − π ) − π f ( x ) = π tan ( π x − π ) − π

f ( x ) = 2 csc ( x ) f ( x ) = 2 csc ( x )

f ( x ) = − 1 4 csc ( x ) f ( x ) = − 1 4 csc ( x )

f ( x ) = 4 sec ( 3 x ) f ( x ) = 4 sec ( 3 x )

f ( x ) = − 3 cot ( 2 x ) f ( x ) = − 3 cot ( 2 x )

f ( x ) = 7 sec ( 5 x ) f ( x ) = 7 sec ( 5 x )

f ( x ) = 9 10 csc ( π x ) f ( x ) = 9 10 csc ( π x )

f ( x ) = 2 csc ( x + π 4 ) − 1 f ( x ) = 2 csc ( x + π 4 ) − 1

f ( x ) = − sec ( x − π 3 ) − 2 f ( x ) = − sec ( x − π 3 ) − 2

f ( x ) = 7 5 csc ( x − π 4 ) f ( x ) = 7 5 csc ( x − π 4 )

f ( x ) = 5 ( cot ( x + π 2 ) − 3 ) f ( x ) = 5 ( cot ( x + π 2 ) − 3 )

For the following exercises, find and graph two periods of the periodic function with the given stretching factor, | A | , | A | , period, and phase shift.

A tangent curve, A = 1 , A = 1 , period of π 3 ; π 3 ; and phase shift ( h , k ) = ( π 4 , 2 ) ( h , k ) = ( π 4 , 2 )

A tangent curve, A = −2 , A = −2 , period of π 4 , π 4 , and phase shift ( h , k ) = ( − π 4 , −2 ) ( h , k ) = ( − π 4 , −2 )

For the following exercises, find an equation for the graph of each function.

For the following exercises, use a graphing calculator to graph two periods of the given function. Note: most graphing calculators do not have a cosecant button; therefore, you will need to input csc x csc x as 1 sin x . 1 sin x .

f ( x ) = | csc ( x ) | f ( x ) = | csc ( x ) |

f ( x ) = | cot ( x ) | f ( x ) = | cot ( x ) |

f ( x ) = csc ( x ) sec ( x ) f ( x ) = csc ( x ) sec ( x )

Graph f ( x ) = 1 + sec 2 ( x ) − tan 2 ( x ) . f ( x ) = 1 + sec 2 ( x ) − tan 2 ( x ) . What is the function shown in the graph?

f ( x ) = sec ( 0.001 x ) f ( x ) = sec ( 0.001 x )

f ( x ) = cot ( 100 π x ) f ( x ) = cot ( 100 π x )

f ( x ) = sin 2 x + cos 2 x f ( x ) = sin 2 x + cos 2 x

Real-World Applications

The function f ( x ) = 20 tan ( π 10 x ) f ( x ) = 20 tan ( π 10 x ) marks the distance in the movement of a light beam from a police car across a wall for time x , x , in seconds, and distance f ( x ) , f ( x ) , in feet.

  • ⓐ Graph on the interval [ 0 , 5 ] . [ 0 , 5 ] .
  • ⓑ Find and interpret the stretching factor, period, and asymptote.
  • ⓒ Evaluate f ( 1 ) f ( 1 ) and f ( 2.5 ) f ( 2.5 ) and discuss the function’s values at those inputs.

Standing on the shore of a lake, a fisherman sights a boat far in the distance to his left. Let x , x , measured in radians, be the angle formed by the line of sight to the ship and a line due north from his position. Assume due north is 0 and x x is measured negative to the left and positive to the right. (See Figure 19 .) The boat travels from due west to due east and, ignoring the curvature of the Earth, the distance d ( x ) , d ( x ) , in kilometers, from the fisherman to the boat is given by the function d ( x ) = 1.5 sec ( x ) . d ( x ) = 1.5 sec ( x ) .

  • ⓐ What is a reasonable domain for d ( x ) ? d ( x ) ?
  • ⓑ Graph d ( x ) d ( x ) on this domain.
  • ⓒ Find and discuss the meaning of any vertical asymptotes on the graph of d ( x ) . d ( x ) .
  • ⓓ Calculate and interpret d ( − π 3 ) . d ( − π 3 ) . Round to the second decimal place.
  • ⓔ Calculate and interpret d ( π 6 ) . d ( π 6 ) . Round to the second decimal place.
  • ⓕ What is the minimum distance between the fisherman and the boat? When does this occur?

A laser rangefinder is locked on a comet approaching Earth. The distance g ( x ) , g ( x ) , in kilometers, of the comet after x x days, for x x in the interval 0 to 30 days, is given by g ( x ) = 250,000 csc ( π 30 x ) . g ( x ) = 250,000 csc ( π 30 x ) .

  • ⓐ Graph g ( x ) g ( x ) on the interval [ 0 , 30 ] . [ 0 , 30 ] .
  • ⓑ Evaluate g ( 5 ) g ( 5 ) and interpret the information.
  • ⓒ What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond?
  • ⓓ Find and discuss the meaning of any vertical asymptotes.

A video camera is focused on a rocket on a launching pad 2 miles from the camera. The angle of elevation from the ground to the rocket after x x seconds is π 120 x . π 120 x .

  • ⓐ Write a function expressing the altitude h ( x ) , h ( x ) , in miles, of the rocket above the ground after x x seconds. Ignore the curvature of the Earth.
  • ⓑ Graph h ( x ) h ( x ) on the interval ( 0 , 60 ) . ( 0 , 60 ) .
  • ⓒ Evaluate and interpret the values h ( 0 ) h ( 0 ) and h ( 30 ) . h ( 30 ) .
  • ⓓ What happens to the values of h ( x ) h ( x ) as x x approaches 60 seconds? Interpret the meaning of this in terms of the problem.

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  • Authors: Jay Abramson
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  • Book title: Precalculus 2e
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  • Book URL: https://openstax.org/books/precalculus-2e/pages/1-introduction-to-functions
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Chapter 7.3: Graphs of the Other Trigonometric Functions

Learning objectives.

In this section, you will:

  • Analyze the graph of  y=tan x.
  • Graph variations of  y=tan x.
  • Analyze the graphs of  y=sec x  and  y=csc x.
  • Graph variations of  y=sec x  and  y=csc x.
  • Analyze the graph of  y=cot x.
  • Graph variations of  y=cot x.

We know the tangent function can be used to find distances, such as the height of a building, mountain, or flagpole. But what if we want to measure repeated occurrences of distance? Imagine, for example, a police car parked next to a warehouse. The rotating light from the police car would travel across the wall of the warehouse in regular intervals. If the input is time, the output would be the distance the beam of light travels. The beam of light would repeat the distance at regular intervals. The tangent function can be used to approximate this distance. Asymptotes would be needed to illustrate the repeated cycles when the beam runs parallel to the wall because, seemingly, the beam of light could appear to extend forever. The graph of the tangent function would clearly illustrate the repeated intervals. In this section, we will explore the graphs of the tangent and other trigonometric functions.

Analyzing the Graph of y = tan x

We will begin with the graph of the tangent function, plotting points as we did for the sine and cosine functions. Recall that

\mathrm{tan}\,x=\frac{\mathrm{sin}\,x}{\mathrm{cos}\,x}

We can determine whether tangent is an odd or even function by using the definition of tangent.

\begin{array}{ll}\mathrm{tan}\left(-x\right)=\frac{\mathrm{sin}\left(-x\right)}{\mathrm{cos}\left(-x\right)}\hfill & \begin{array}{ccc}& & \end{array}\text{Definition of tangent}.\hfill \\ \text{ }=\frac{-\mathrm{sin}\,x}{\mathrm{cos}\,x}\hfill & \begin{array}{ccc}& & \end{array}\text{Sine is an odd function, cosine is even}.\hfill \\ \text{ }=-\frac{\mathrm{sin}\,x}{\mathrm{cos}\,x}\hfill & \begin{array}{ccc}& & \end{array}\text{The quotient of an odd and an even function is odd}.\hfill \\ \text{ }=-\mathrm{tan}\,x\hfill & \begin{array}{ccc}& & \end{array}\text{Definition of tangent}.\hfill \end{array}

Therefore, tangent is an odd function. We can further analyze the graphical behavior of the tangent function by looking at values for some of the special angles, as listed in (Figure) .

\,\frac{\pi }{3}<x<\frac{\pi }{2},\,

Graphing Variations of y = tan x

As with the sine and cosine functions, the tangent function can be described by a general equation.

y=A\mathrm{tan}\left(Bx\right)

Features of the Graph of y = A tan( Bx )

\,|A|.

Graphing One Period of a Stretched or Compressed Tangent Function

\,f\left(x\right)=A\mathrm{tan}\left(Bx\right).\,

Sketching a Compressed Tangent

\,y=0.5\mathrm{tan}\left(\frac{\pi }{2}x\right).

Graphing One Period of a Shifted Tangent Function

\,C\,

Features of the Graph of y = A tan( Bx − C )+ D

\,\frac{\pi }{|B|}.

  • There is no amplitude.

y=A\,\mathrm{tan}\left(Bx-C\right)+D\,

  • Plot any three reference points and draw the graph through these points.

\,y=-2\mathrm{tan}\left(\pi x+\pi \right)\,-1.

Given the graph of a tangent function, identify horizontal and vertical stretches.

\,P\,

Identifying the Graph of a Stretched Tangent

Find a formula for the function graphed in (Figure) .

A graph of two periods of a modified tangent function, with asymptotes at x=-4 and x=4.

The graph has the shape of a tangent function.

\,P=8.\,

Find a formula for the function in (Figure) .

A graph of four periods of a modified tangent function, Vertical asymptotes at -3pi/4, -pi/4, pi/4, and 3pi/4.

Analyzing the Graphs of y = sec x and y = csc x

\,\mathrm{sec}\,x=\frac{1}{\mathrm{cos}\,x}.\,

Features of the Graph of y = A sec( Bx )

\,\frac{2\pi }{|B|}.

The graph of cosecant, which is shown in (Figure) , is similar to the graph of secant.

A graph of cosecant of x and sin of x. Five vertical asymptotes shown at multiples of pi.

Features of the Graph of y = A csc( Bx )

\,x\ne \frac{\pi }{|B|}k,\,

Graphing Variations of y = sec x and y = csc x

For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function in the same way as for the secant and other functions.The equations become the following.

y=A\mathrm{sec}\left(Bx-C\right)+D

Features of the Graph of y = A sec( Bx − C )+ D

\,x\ne \frac{C}{B}+\frac{\pi }{2|B|}k,

Features of the Graph of y = A csc( Bx − C )+ D

\,x=\frac{C}{B}+\frac{\pi }{|B|}k,

  • Sketch the asymptotes.
  • Plot any two reference points and draw the graph through these points.

Graphing a Variation of the Secant Function

\,f\left(x\right)=2.5\mathrm{sec}\left(0.4x\right).

  • Step 5. Use the reciprocal relationship of the cosine and secant functions to draw the cosecant function.

\,x=1.25\pi \,

Do the vertical shift and stretch/compression affect the secant’s range?

\,f\left(x\right)=A\mathrm{sec}\left(Bx-C\right)+D\,

Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original function’s input.

\,y=A\mathrm{csc}\left(Bx\right),\,

Graphing a Variation of the Cosecant Function

\,f\left(x\right)=-3\mathrm{csc}\left(4x\right).

  • Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function .

\,x=0,\,x=\frac{\pi }{4},\,

Graphing a Vertically Stretched, Horizontally Compressed, and Vertically Shifted Cosecant

\,y=2\mathrm{csc}\left(\frac{\pi }{2}x\right)+1.\,

The graph for this function is shown in (Figure) .

A graph of 3 periods of a modified cosecant function, with 3 vertical asymptotes, and a dotted sinusoidal function that has local maximums where the cosecant function has local minimums and local minimums where the cosecant function has local maximums.

Analyzing the Graph of y = cot x

\,\mathrm{cot}\,x=\frac{1}{\mathrm{tan}\,x}.\,

Features of the Graph of y = A cot( Bx )

\,\left(-\infty ,\infty \right).

Graphing Variations of y = cot x

We can transform the graph of the cotangent in much the same way as we did for the tangent. The equation becomes the following.

y=A\mathrm{cot}\left(Bx-C\right)+D

Properties of the Graph of y = A cot( Bx −C)+ D

\,f\left(x\right)=A\mathrm{cot}\left(Bx\right),

  • Plot any two reference points.

\,y=A\mathrm{cot}\left(Bx\right).

Graphing Variations of the Cotangent Function

\,y=3\mathrm{cot}\left(4x\right),\,

Graphing a Modified Cotangent

\,f\left(x\right)=4\mathrm{cot}\left(\frac{\pi }{8}x-\frac{\pi }{2}\right)-2.

The graph is shown in (Figure) .

A graph of one period of a modified cotangent function. Vertical asymptotes at x=4 and x=12.

Using the Graphs of Trigonometric Functions to Solve Real-World Problems

Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function .

Using Trigonometric Functions to Solve Real-World Scenarios

\,y=5\mathrm{tan}\left(\frac{\pi }{4}t\right)\,

  • Find and interpret the stretching factor and period.

\,\left[0,5\right].

We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period.

\,\frac{\pi }{\frac{\pi }{4}}=\frac{\pi }{1}\cdot \frac{4}{\pi }=4.\,

Access these online resources for additional instruction and practice with graphs of other trigonometric functions.

  • Graphing the Tangent
  • Graphing Cosecant and Secant
  • Graphing the Cotangent

Key Equations

Key concepts.

\,\pi .

  • Real-world scenarios can be solved using graphs of trigonometric functions. See (Figure) .

Section Exercises

\,y=\mathrm{csc}\,x.

For the following exercises, match each trigonometric function with one of the following graphs.

Trigonometric graph of tangent of x.

For the following exercises, find the period and horizontal shift of each of the functions.

f\left(x\right)=2\mathrm{tan}\left(4x-32\right)

period: 8; horizontal shift: 1 unit to left

m\left(x\right)=6\mathrm{csc}\left(\frac{\pi }{3}x+\pi \right)

For the following exercises, sketch two periods of the graph for each of the following functions. Identify the stretching factor, period, and asymptotes.

A graph of two periods of a modified tangent function. There are two vertical asymptotes.

For the following exercises, find an equation for the graph of each function.

A graph of two periods of a modified cosecant function, with asymptotes at multiples of pi/2.

Real-World Applications

\,f\left(x\right)=20\mathrm{tan}\left(\frac{\pi }{10}x\right)\,

  • Find and interpret the stretching factor, period, and asymptote.

\,f\left(2.5\right)\,

  • What is the minimum distance between the fisherman and the boat? When does this occur?

An illustration of a man and the distance he is away from a boat.

  • What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond?
  • Find and discuss the meaning of any vertical asymptotes.

\,\frac{\pi }{120}x.

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Graphs of Trigonometric Functions - Problem Solving

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To solve the problems on this page, you should be familiar with the following:

  • Sine and Cosine Graphs
  • Cosec and Sec Graphs
  • Tangent and Cotangent Graphs
  • Inverse Trigonometric Graphs
  • Trigonometric Graphs - Amplitude and Periodicity
  • Graphical Transformation of Trigonometric Functions

Problem Solving - Basic

Problem solving - intermediate, problem solving - advanced.

What is the fundamental period of the function \(f(x) = \cos^2\big(\sin(x)\big)? \) Note that by graphing, we can see that \( \cos^2\big(\sin(x)\big) = \cos^2\big(\sin(x+ k\pi)\big) \) holds true for all integer \(k\). Thus the fundamental period of \(f(x) \) is \(\pi\). \(_\square\) P.S. We can also solve this via compound angle formula.
Evaluate \(\sin \big(10^{10}\big)^\circ \div \sin 80^\circ.\) Because \(y = \sin x \) has a fundamental period of 360 degrees, \[\sin \big(10^{10}\big)^\circ = \sin \big(10^{10} \bmod{360}\big)^\circ = \sin 280^\circ = -\sin 80^\circ. \] Thus taking the quotient yields \(-1\) as the answer. \(_\square\)

\[ \large\tan(x)+\sec(x)=2\cos(x)\]

Find the number of solutions of \(x\) in the interval \([0,2\pi] \) that satisfy the equation above.

\[ \large \color{purple}{\sin^{-1}} \left [ \color{blue} {\sin} (\color{green}{10}) \right ] = \, \color{brown}? \]

Given that \(\tan 1^\circ > \frac1{90},\) which of these numbers is larger, \(\tan(\tan 1^\circ)\) or \(\tan(\cot 1^\circ)?\) Because \(\cot 1^\circ \) is the reciprocal of \(\tan 1^\circ, \) which is rather small, \( \tan 1^\circ < \cot 1^\circ.\) Also, because both of them are positive and less than 90, they are in the first quadrant. With \(y = \tan x \) as an increasing function, we have \( \tan( \tan 1^\circ) < \tan(\cot 1^\circ). \) So the latter number is larger. \(_\square \)

For \(x\in\mathbb{Z}\), find the probability that

\[2\sin x^{\circ}<1.\]

\[ A = \max_{x \in \mathbb{R}} \left( \log_2 3 \right)^{\sin x }, \qquad B = \max_{x \in \mathbb{R}} \left( \log_3 2 \right)^{\sin x }\]

Which is larger, \(A\) or \(B?\)

\[\large {f(x)= \displaystyle \lim_{n \to \infty} \dfrac{x}{(2\sin x)^{2n}+1}}\]

How many values of \(x\) are there from \(0\) to \(\frac{9\pi}{2}\) (both inclusive) such that \(f(x)\) is discontinuous at those values of \(x\).

\[ A = \sin\left[\sin(1)\right] \\ B = \sin\left[\cos(1)\right] \\ C = \cos\left[\sin(1)\right] \\ D =\cos\left[\cos(1)\right] \]

The above are the values of \(A,B,C,\) and \(D.\) Which of the answer choices is true?

\(\) Clarification: All angles are measured in radians.

Inspiration.

Find the total number of solutions of the equation

\[2x=3\pi(1-\cos x).\]

\[\large \prod_{r = 1}^{12} \sin (rx) = 0\]

What is the number of solutions of \(x\) satisfying the equation above in the interval \((0,\pi]?\)

How many real numbers \(x\) satisfy \(\sin x = \frac{x}{100}?\)

For every integer \(k,\) we define a function \(f_k \) by the formula

\[f_k(x)=100x-k\sin x.\]

What is the smallest positive integer value of \(k\) such that, for some real \(\alpha\), we have \(f_k\big(f_k(\alpha)\big)=\alpha,\) but \(f_k(\alpha )\neq \alpha?\)

\(\) Details and Assumptions: The function is evaluated in radians. There is no degree symbol in the problem.

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Unit 2: Trigonometric functions

About this unit, unit circle introduction.

  • Unit circle (Opens a modal)
  • The trig functions & right triangle trig ratios (Opens a modal)
  • Trig unit circle review (Opens a modal)
  • Trigonometric functions: FAQ (Opens a modal)
  • Unit circle Get 3 of 4 questions to level up!
  • Intro to radians (Opens a modal)
  • Radians & degrees (Opens a modal)
  • Degrees to radians (Opens a modal)
  • Radians to degrees (Opens a modal)
  • Radian angles & quadrants (Opens a modal)
  • Radians & degrees Get 3 of 4 questions to level up!
  • Unit circle (with radians) Get 3 of 4 questions to level up!

The Pythagorean identity

  • Proof of the Pythagorean trig identity (Opens a modal)
  • Using the Pythagorean trig identity (Opens a modal)
  • Pythagorean identity review (Opens a modal)
  • Use the Pythagorean identity Get 3 of 4 questions to level up!

Special trigonometric values in the first quadrant

  • Cosine, sine and tangent of π/6 and π/3 (Opens a modal)
  • Trig values of π/4 (Opens a modal)
  • Trig values of π/6, π/4, and π/3 Get 3 of 4 questions to level up!

Trigonometric values on the unit circle

  • Sine & cosine identities: symmetry (Opens a modal)
  • Tangent identities: symmetry (Opens a modal)
  • Sine & cosine identities: periodicity (Opens a modal)
  • Tangent identities: periodicity (Opens a modal)
  • Trig identities from reflections and rotations Get 3 of 4 questions to level up!
  • Trig values of special angles Get 3 of 4 questions to level up!

Graphs of sin(x), cos(x), and tan(x)

  • Graph of y=sin(x) (Opens a modal)
  • Intersection points of y=sin(x) and y=cos(x) (Opens a modal)
  • Graph of y=tan(x) (Opens a modal)

Amplitude, midline, and period

  • Features of sinusoidal functions (Opens a modal)
  • Midline, amplitude, and period review (Opens a modal)
  • Midline of sinusoidal functions from graph Get 3 of 4 questions to level up!
  • Amplitude of sinusoidal functions from graph Get 3 of 4 questions to level up!
  • Period of sinusoidal functions from graph Get 3 of 4 questions to level up!

Transforming sinusoidal graphs

  • Amplitude & period of sinusoidal functions from equation (Opens a modal)
  • Transforming sinusoidal graphs: vertical stretch & horizontal reflection (Opens a modal)
  • Transforming sinusoidal graphs: vertical & horizontal stretches (Opens a modal)
  • Amplitude of sinusoidal functions from equation Get 3 of 4 questions to level up!
  • Midline of sinusoidal functions from equation Get 3 of 4 questions to level up!
  • Period of sinusoidal functions from equation Get 3 of 4 questions to level up!

Graphing sinusoidal functions

  • Example: Graphing y=3⋅sin(½⋅x)-2 (Opens a modal)
  • Example: Graphing y=-cos(π⋅x)+1.5 (Opens a modal)
  • Sinusoidal function from graph (Opens a modal)
  • Graph sinusoidal functions Get 3 of 4 questions to level up!
  • Construct sinusoidal functions Get 3 of 4 questions to level up!
  • Graph sinusoidal functions: phase shift Get 3 of 4 questions to level up!

Sinusoidal models

  • Interpreting trigonometric graphs in context (Opens a modal)
  • Trig word problem: modeling daily temperature (Opens a modal)
  • Trig word problem: modeling annual temperature (Opens a modal)
  • Trig word problem: length of day (phase shift) (Opens a modal)
  • Interpreting trigonometric graphs in context Get 3 of 4 questions to level up!
  • Modeling with sinusoidal functions Get 3 of 4 questions to level up!
  • Modeling with sinusoidal functions: phase shift Get 3 of 4 questions to level up!

Long live Tau

  • Tau versus pi (Opens a modal)
  • Pi is (still) wrong (Opens a modal)

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Solving Real-Life Problems Using Trigonometry

Trigonometry - Practical Problems - Worksheet B

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Periodic Functions

Graphs of the other trigonometric functions, learning outcomes.

  • Analyze the graph of  y=tan x and y=cot x.
  • Graph variations of  y=tan x and y=cot x.
  • Determine a function formula from a tangent or cotangent graph.
  • Analyze the graphs of  y=sec x  and  y=csc x.
  • Graph variations of  y=sec x  and  y=csc x.
  • Determine a function formula from a secant or cosecant graph.

Analyzing the Graph of y = tan x and Its Variations

We will begin with the graph of the tangent function, plotting points as we did for the sine and cosine functions. Recall that

The period of the tangent function is π because the graph repeats itself on intervals of kπ where k is a constant. If we graph the tangent function on [latex]−\dfrac{\pi}{2}\text{ to }\dfrac{\pi}{2}[/latex], we can see the behavior of the graph on one complete cycle. If we look at any larger interval, we will see that the characteristics of the graph repeat.

We can determine whether tangent is an odd or even function by using the definition of tangent.

[latex]\begin{align}\tan(−x)&=\frac{\sin(−x)}{\cos(−x)} && \text{Definition of tangent.} \\ &=\frac{−\sin x}{\cos x} && \text{Sine is an odd function, cosine is even.} \\ &=−\frac{\sin x}{\cos x} && \text{The quotient of an odd and an even function is odd.} \\ &=−\tan x && \text{Definition of tangent.} \end{align}[/latex]

Therefore, tangent is an odd function. We can further analyze the graphical behavior of the tangent function by looking at values for some of the special angles, as listed in the table below.

These points will help us draw our graph, but we need to determine how the graph behaves where it is undefined. If we look more closely at values when [latex]\frac{\pi}{3}<x<\frac{\pi}{2}[/latex], we can use a table to look for a trend. Because [latex]\frac{\pi}{3}\approx 1.05[/latex] and [latex]\frac{\pi}{2}\approx 1.57[/latex], we will evaluate x at radian measures 1.05 < x < 1.57 as shown in the table below.

As x approaches [latex]\frac{\pi}{2}[/latex], the outputs of the function get larger and larger. Because [latex]y=\tan x[/latex] is an odd function, we see the corresponding table of negative values in the table below.

We can see that, as x approaches [latex]−\dfrac{\pi}{2}[/latex], the outputs get smaller and smaller. Remember that there are some values of x for which cos x = 0. For example, [latex]\cos\left(\frac{\pi}{2}\right)=0[/latex] and [latex]\cos\left(\frac{3\pi}{2}\right)=0[/latex]. At these values, the tangent function is undefined, so the graph of [latex]y=\tan x[/latex] has discontinuities at [latex]x=\frac{\pi}{2}[/latex] and [latex]\frac{3\pi}{2}[/latex]. At these values, the graph of the tangent has vertical asymptotes. Figure 1 represents the graph of [latex]y=\tan x[/latex]. The tangent is positive from 0 to [latex]\frac{\pi}{2}[/latex] and from π to [latex]\frac{3\pi}{2}[/latex], corresponding to quadrants I and III of the unit circle.

A graph of y=tangent of x. Asymptotes at -pi over 2 and pi over 2.

Figure 1. Graph of the tangent function

Graphing Variations of y = tan x

As with the sine and cosine functions, the tangent function can be described by a general equation.

We can identify horizontal and vertical stretches and compressions using values of A and B. The horizontal stretch can typically be determined from the period of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph.

Because there are no maximum or minimum values of a tangent function, the term amplitude cannot be interpreted as it is for the sine and cosine functions. Instead, we will use the phrase stretching/compressing factor when referring to the constant A.

A General Note: Features of the Graph of y = A tan( Bx )

  • The stretching factor is | A | .
  • The period is [latex]P=\frac{\pi}{|B|}[/latex].
  • The domain is all real numbers x , where [latex]x\ne \frac{\pi}{2|B|} + \frac{\pi}{|B|} k[/latex] such that k is an integer.
  • The range is [latex]\left(-\infty,\infty\right)[/latex].
  • The asymptotes occur at [latex]x=\frac{\pi}{2|B|} + \frac{\pi}{|B|}k[/latex], where k is an integer.
  • [latex]y = A \tan (Bx)[/latex] is an odd function.

Graphing One Period of a Stretched or Compressed Tangent Function

We can use what we know about the properties of the tangent function to quickly sketch a graph of any stretched and/or compressed tangent function of the form [latex]f(x)=A\tan(Bx)[/latex]. We focus on a single period of the function including the origin, because the periodic property enables us to extend the graph to the rest of the function’s domain if we wish. Our limited domain is then the interval [latex](−\frac{P}{2}, \frac{P}{2})[/latex] and the graph has vertical asymptotes at [latex]\pm \frac{P}{2}[/latex] where [latex]P=\frac{\pi}{B}[/latex]. On [latex](−\dfrac{\pi}{2}, \dfrac{\pi}{2})[/latex], the graph will come up from the left asymptote at [latex]x=−\dfrac{\pi}{2}[/latex], cross through the origin, and continue to increase as it approaches the right asymptote at [latex]x=\frac{\pi}{2}[/latex]. To make the function approach the asymptotes at the correct rate, we also need to set the vertical scale by actually evaluating the function for at least one point that the graph will pass through. For example, we can use

because  [latex]\tan\left(\frac{\pi}{4}\right)=1[/latex].

How To: Given the function [latex]f(x)=A\tan(Bx)[/latex], graph one period.

  • Identify the stretching factor, |A|.
  • Identify B and determine the period, [latex]P=\frac{\pi}{|B|}[/latex].
  • Draw vertical asymptotes at  [latex]x=−\dfrac{P}{2}[/latex] and [latex]x=\frac{P}{2}[/latex].
  • For A > 0 , the graph approaches the left asymptote at negative output values and the right asymptote at positive output values (reverse for A < 0 ).
  • Plot reference points at [latex]\left(\frac{P}{4},A\right)[/latex] (0, 0), and ([latex]−\dfrac{P}{4}[/latex],− A), and draw the graph through these points.

Example 1: Sketching a Compressed Tangent

Sketch a graph of one period of the function [latex]y=0.5\tan\left(\frac{\pi}{2}x\right)[/latex].

First, we identify A and B.

An illustration of equations showing that A is the coefficient of tangent and B is the coefficient of x, which is within the tangent function.

Because [latex]A=0.5[/latex] and [latex]B=\frac{\pi}{2}[/latex], we can find the stretching/compressing factor and period. The period is [latex]\frac{\pi}{\frac{\pi}{2}}=2[/latex], so the asymptotes are at [latex]x=\pm 1[/latex]. At a quarter period from the origin, we have

[latex]\begin{align}f(0.5)&=0.5\tan\left(\frac{0.5\pi}{2}\right)\\ &=0.5\tan(\frac{\pi}{4})\\ &=0.5 \end{align}[/latex]

This means the curve must pass through the points(0.5,0.5),(0,0),and(−0.5,−0.5).The only inflection point is at the origin. Figure shows the graph of one period of the function.

A graph of one period of a modified tangent function, with asymptotes at x=-1 and x=1.

Sketch a graph of [latex]f(x)=3\tan\left(\frac{\pi}{6}x\right)[/latex].

A graph of two periods of a modified tangent function, with asymptotes at x=-3 and x=3.

Graphing One Period of a Shifted Tangent Function

Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we add C and D to the general form of the tangent function.

The graph of a transformed tangent function is different from the basic tangent function tan x in several ways:

A General Note: Features of the Graph of [latex]y = A\tan\left(Bx−C\right)+D[/latex]

  • The stretching factor is | A |.
  • The period is [latex]\frac{\pi}{|B|}[/latex].
  • The domain is [latex]x\ne\frac{C}{B}+\frac{\pi}{|B|}k[/latex], where k is an integer.
  • The range is (−∞,−| A |] ∪ [| A |, ∞).
  • The vertical asymptotes occur at [latex]x=\frac{C}{B}+\frac{\pi}{2|B|}k[/latex], where k is an odd integer.
  • There is no amplitude.
  • [latex]y=A\tan(Bx)[/latex] is an odd function because it is the quotient of odd and even functions (sine and cosine respectively).

How To: Given the function [latex]y=A\tan(Bx−C)+D[/latex], sketch the graph of one period.

  • Express the function given in the form [latex]y=A\tan(Bx−C)+D[/latex].
  • Identify the stretching/compressing factor, |A|.
  • Identify C and determine the phase shift, [latex]\frac{C}{B}[/latex].
  • Draw the graph of [latex]y=A\tan(Bx)[/latex] shifted to the right by [latex]\frac{C}{B}[/latex] and up by D .
  • Sketch the vertical asymptotes, which occur at [latex]x=\frac{C}{B}+\frac{\pi}{2|B|}k[/latex], where k is an odd integer.
  • Plot any three reference points and draw the graph through these points.

Example 2: Graphing One Period of a Shifted Tangent Function

Graph one period of the function [latex]y=−2\tan(\pi x+\pi)−1[/latex].

Step 1. The function is already written in the form [latex]y=A\tan(Bx−C)+D[/latex].

Step 2.  [latex]A=−2[/latex], so the stretching factor is [latex]|A|=2[/latex].

Step 3.  [latex]B=\pi[/latex], so the period is [latex]P=\frac{\pi}{|B|}=\frac{\pi}{\pi}=1[/latex].

Step 4.  [latex]C=−\pi[/latex], so the phase shift is [latex]\dfrac{C}{B}=\dfrac{−\pi}{\pi}=−1[/latex].

Step 5–7. The asymptotes are at [latex]x=−\frac{3}{2}[/latex] and [latex]x=−\frac{1}{2}[/latex] and the three recommended reference points are (−1.25, 1), (−1,−1), and (−0.75, −3). The graph is shown in Figure 4.

A graph of one period of a shifted tangent function, with vertical asymptotes at x=-1.5 and x=-0.5.

Analysis of the Solution

Note that this is a decreasing function because A < 0.

How would the graph in Example 2 look different if we made A = 2 instead of −2?

It would be reflected across the line [latex]y=−1[/latex], becoming an increasing function.

How To: Given the graph of a tangent function, identify horizontal and vertical stretches.

  • Find the period P from the spacing between successive vertical asymptotes or x -intercepts.
  • Write [latex]f(x)=A\tan\left(\frac{\pi}{P}x\right)[/latex].
  • Determine a convenient point ( x , f ( x )) on the given graph and use it to determine A .

Example 3: Identifying the Graph of a Stretched Tangent

Find a formula for the function graphed in Figure 5.

A graph of two periods of a modified tangent function, with asymptotes at x=-4 and x=4.

The graph has the shape of a tangent function.

Step 1. One cycle extends from –4 to 4, so the period is [latex]P=8[/latex]. Since [latex]P=\frac{\pi}{|B|}[/latex], we have [latex]B=\frac{\pi}{P}=\frac{\pi}{8}[/latex].

Step 2. The equation must have the [latex]\text{form}f(x)=A\tan\left(\frac{\pi}{8}x\right)[/latex].

Step 3. To find the vertical stretch A , we can use the point (2,2).

[latex]2=A\tan\left(\frac{\pi}{8}\times2\right)=A\tan\left(\frac{\pi}{4}\right)[/latex]

Because [latex]\tan\left(\frac{\pi}{4}\right)=1[/latex], A = 2.

This function would have a formula [latex]f(x)=2\tan\left(\frac{\pi}{8}x\right)[/latex].

Find a formula for the function in Figure 6.

A graph of four periods of a modified tangent function, Vertical asymptotes at -3pi/4, -pi/4, pi/4, and 3pi/4.

[latex]g(x)=4\tan(2x)[/latex]

Using the Graphs of Trigonometric Functions to Solve Real-World Problems

Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function .

Example 4: Using Trigonometric Functions to Solve Real-World Scenarios

Suppose the function [latex]y=5\tan\left(\frac{\pi}{4}t\right)[/latex] marks the distance in the movement of a light beam from the top of a police car across a wall where t is the time in seconds and y is the distance in feet from a point on the wall directly across from the police car.

  • Find and interpret the stretching factor and period.
  • Graph on the interval [0, 5].
  • Evaluate f (1) and discuss the function’s value at that input.

A graph showing that variable A is the coefficient of the tangent function and variable B is the coefficient of x, which is within that tangent function.

We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period.

A graph of one period of a modified tangent function, with a vertical asymptote at x=4.

  • period: [latex]f(1)=5\tan \left(\frac{\pi}{4}\left(1\right)\right)=5\left(1\right)=5[/latex]; after 1 second, the beam of has moved 5 ft from the spot across from the police car.

Analyzing the Graphs of y = sec x and y = cscx and Their Variations

The secant was defined by the reciprocal identity  [latex]\sec x=\frac{1}{\cos x}[/latex]. Notice that the function is undefined when the cosine is 0, leading to vertical asymptotes at [latex]\frac{\pi}{2},\frac{3\pi}{2}\text{, etc}[/latex]. Because the cosine is never more than 1 in absolute value, the secant, being the reciprocal, will never be less than 1 in absolute value.

We can graph [latex]y=\sec x[/latex] by observing the graph of the cosine function because these two functions are reciprocals of one another. See Figure 9. The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined.

The secant graph has vertical asymptotes at each value of x where the cosine graph crosses the x -axis; we show these in the graph below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant.

Note that, because cosine is an even function, secant is also an even function. That is, [latex]\sec(−x)=\sec x[/latex].

A graph of cosine of x and secant of x. Asymptotes for secant of x shown at -3pi/2, -pi/2, pi/2, and 3pi/2.

As we did for the tangent function, we will again refer to the constant | A | as the stretching factor, not the amplitude.

A General Note: Features of the Graph of y = A sec( Bx )

  • The period is [latex]\frac{2\pi}{|B|}[/latex].
  • The domain is [latex]x\ne \frac{\pi}{2|B|}k[/latex], where k is an odd integer.
  • The range is (−∞, −| A |] ∪ [| A |, ∞).
  • The vertical asymptotes occur at [latex]x=\frac{\pi}{2|B|}k [/latex], where k is an odd integer.
  • [latex]y=A\sec(Bx)[/latex] is an even function because cosine is an even function.

Similar to the secant, the cosecant is defined by the reciprocal identity [latex]\csc x=1\sin x[/latex]. Notice that the function is undefined when the sine is 0, leading to a vertical asymptote in the graph at 0, π, etc. Since the sine is never more than 1 in absolute value, the cosecant, being the reciprocal, will never be less than 1 in absolute value.

We can graph [latex]y=\csc x[/latex] by observing the graph of the sine function because these two functions are reciprocals of one another. See Figure 10. The graph of sine is shown as a dashed orange wave so we can see the relationship. Where the graph of the sine function decreases, the graph of the cosecant function increases. Where the graph of the sine function increases, the graph of the cosecant function decreases.

The cosecant graph has vertical asymptotes at each value of x where the sine graph crosses the x -axis; we show these in the graph below with dashed vertical lines.

Note that, since sine is an odd function, the cosecant function is also an odd function. That is, [latex]\csc(−x)=−\csc x[/latex].

The graph of cosecant, which is shown in Figure 10, is similar to the graph of secant.

A graph of cosecant of x and sin of x. Five vertical asymptotes shown at multiples of pi.

A General Note: Features of the Graph of [latex]y=A\csc(Bx)

  • The domain is [latex]x\ne\frac{\pi}{|B|}k[/latex], where k is an integer.
  • The range is ( −∞, −|A|] ∪ [|A|, ∞).
  • The asymptotes occur at [latex]x=\frac{\pi}{|B|}k[/latex], where k is an integer.
  • [latex]y=A\csc(Bx)[/latex] is an odd function because sine is an odd function.

Graphing Variations of y = sec x and y  = csc x

For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function in the same way as for the secant and other functions. The equations become the following.

A General Note: Features of the Graph of [latex]y=A\sec(Bx−C)+D[/latex]

  • The domain is [latex]x\ne \frac{C}{B}+\frac{\pi}{2|B|}k[/latex], where k is an odd integer.

A General Note: Features of the Graph of [latex]y=A\csc(Bx−C)+D[/latex]

  • The domain is [latex]x\ne\frac{C}{B}+\frac{\pi}{2|B|}k[/latex], where k is an integer.
  • The vertical asymptotes occur at [latex]x=\frac{C}{B}+\frac{\pi}{|B|}k[/latex], where k is an integer.

How To: Given a function of the form [latex]y=A\sec(Bx)[/latex], graph one period.

  • Express the function given in the form [latex]y=A\sec(Bx)[/latex].
  • Identify B and determine the period, [latex]P=\frac{2\pi}{|B|}[/latex].
  • Sketch the graph of [latex]y=A\cos(Bx)[/latex].
  • Use the reciprocal relationship between [latex]y=\cos x[/latex] and [latex]y=\sec x[/latex] to draw the graph of [latex]y=A\sec(Bx)[/latex].
  • Sketch the asymptotes.
  • Plot any two reference points and draw the graph through these points.

Example 6: Graphing a Variation of the Secant Function

Graph one period of [latex]f(x)=2.5\sec(0.4x)[/latex].

Step 1. The given function is already written in the general form, [latex]y=A\sec(Bx)[/latex]. Step 2.  [latex]A=2.5[/latex] so the stretching factor is 2.5. Step 3.  [latex]B=0.4[/latex], so [latex]P=\frac{2\pi}{0.4}=5\pi[/latex]. The period is 5π units. Step 4. Sketch the graph of the function [latex]g(x)=2.5\cos(0.4x)[/latex]. Step 5. Use the reciprocal relationship of the cosine and secant functions to draw the cosecant function. Steps 6–7. Sketch two asymptotes at [latex]x=1.25\pi[/latex] and [latex]x=3.75\pi[/latex]. We can use two reference points, the local minimum at (0, 2.5) and the local maximum at (2.5π, −2.5). Figure 11 shows the graph.

A graph of one period of a modified secant function, which looks like an upward facing prarbola and a downward facing parabola.

Graph one period of [latex]f(x)=−2.5\sec(0.4x)[/latex].

A graph of one period of a modified secant function, which looks like an downward facing prarbola and a upward facing parabola.

Do the vertical shift and stretch/compression affect the secant’s range?

Yes. The range of  [latex]f(x) = A\sec(Bx − C) + D[/latex] is ( −∞, −| A | + D ] ∪ [| A | + D , ∞).

How To: Given a function of the form [latex]f(x)=A\sec (Bx−C)+D[/latex], graph one period.

  • Express the function given in the form [latex]y=A\sec(Bx−C)+D[/latex].
  • Identify the stretching/compressing factor, | A |.
  • Identify B and determine the period, [latex]\frac{2\pi}{|B|}[/latex].
  • Draw the graph of [latex]y=A\sec(Bx)[/latex]. but shift it to the right by [latex]\frac{C}{B}[/latex] and up by D .

Example 7: Graphing a Variation of the Secant Function

Graph one period of [latex]y=4\sec \left(\frac{\pi}{3}x−\frac{\pi}{2}\right)+1[/latex].

Step 1. Express the function given in the form [latex]y=4\sec \left(\frac{\pi}{3}x−\frac{\pi}{2}\right)+1[/latex].

Step 2. The stretching/compressing factor is | A | = 4.

Step 3. The period is

[latex]\begin{align} \frac{2\pi}{|B|}&=\frac{2\pi}{\frac{\pi}{3}}\\ &=\frac{2\pi}{1}\times\frac{3}{\pi}\\ &=6 \end{align}[/latex]

Step 4. The phase shift is

[latex]\begin{align}\frac{C}{B}&=\frac{\frac{\pi}{2}}{\frac{\pi}{3}} \\ &=\frac{\pi}{2} \times \frac{3}{\pi} \\ &=1.5 \end{align}[/latex]

Step 5. Draw the graph of [latex]y=A\sec(Bx)[/latex],but shift it to the right by [latex]\frac{C}{B}=1.5[/latex] and up by D  = 6.

Step 6. Sketch the vertical asymptotes, which occur at x  = 0, x = 3, and x = 6. There is a local minimum at (1.5, 5) and a local maximum at (4.5, −3). Figure 12 shows the graph.

graphing other trigonometric functions solving real world problems

Graph one period of [latex]f(x)=−6\sec(4x+2)−8[/latex].

A graph of one period of a modified secant function. There are two vertical asymptotes, one at approximately x=-pi/20 and one approximately at 3pi/16.

The domain of [latex]\csc x[/latex] was given to be all x such that [latex]x\ne k\pi[/latex] for any integer k . Would the domain of [latex]y=A\csc(Bx−C)+D[/latex] be [latex]x\ne\frac{C+k\pi}{B}[/latex]?

Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original function’s input.

How To: Given a function of the form [latex]y=A\csc(Bx)[/latex], graph one period.

  • Express the function given in the form [latex]y=A\csc(Bx)[/latex].
  • Draw the graph of [latex]y=A\sin(Bx)[/latex].
  • Use the reciprocal relationship between [latex]y=\sin x[/latex] and [latex]y=\csc x[/latex] to draw the graph of [latex]y=A\csc(Bx) [/latex].

Example 8: Graphing a Variation of the Cosecant Function

Graph one period of [latex]f(x)=−3\csc(4x)[/latex].

Step 1. The given function is already written in the general form, [latex]y=A\csc(Bx)[/latex].

Step 2. [latex]|A|=|−3|=3[/latex], so the stretching factor is 3.

Step 3. [latex]B=4\text{, so}P=\frac{2\pi}{4}=\frac{\pi}{2}[/latex].The period is [latex]\frac{\pi}{2}[/latex] units.

Step 4. Sketch the graph of the function [latex]g(x)=−3\sin(4x)[/latex].

Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function.

Steps 6–7. Sketch three asymptotes at [latex]x=0\text{, }x=\frac{\pi}{4}\text{, and }x=\frac{\pi}{2}[/latex].We can use two reference points, the local maximum at [latex]\left(\frac{\pi}{8}\text{, }−3\right)[/latex] and the local minimum at [latex]\left(\frac{3\pi}{8}\text{, }3\right)[/latex]. Figure 13 shows the graph.

A graph of one period of a cosecant function. There are vertical asymptotes at x=0, x=pi/4, and x=pi/2.

Graph one period of [latex]f(x)=0.5\csc(2x)[/latex].

A graph of one period of a modified secant function, which looks like an downward facing prarbola and a upward facing parabola.

How To: Given a function of the form [latex]f(x)=A\csc(Bx−C)+D[/latex], graph one period.

  • Express the function given in the form [latex]y=A\csc(Bx−C)+D[/latex].
  • Draw the graph of [latex]y=A\csc(Bx)[/latex] but shift it to the right by and up by D .
  • Sketch the vertical asymptotes, which occur at [latex]x=\frac{C}{B}+\frac{\pi}{|B|}k[/latex], where k is an integer.

Example 9: Graphing a Vertically Stretched, Horizontally Compressed, and Vertically Shifted Cosecant

Sketch a graph of [latex]y=2\csc\left(\frac{\pi}{2}x\right)+1[/latex]. What are the domain and range of this function?

Step 1. Express the function given in the form [latex]y=2\csc\left(\frac{\pi}{2}x\right)+1[/latex].

Step 2. Identify the stretching/compressing factor, [latex]|A|=2[/latex].

Step 3. The period is [latex]\frac{2\pi}{|B|}=\frac{2\pi}{\frac{\pi}{2}}=\frac{2\pi}{1}\times \frac{2}{\pi}=4[/latex].

Step 4. The phase shift is [latex]\frac{0}{\frac{\pi}{2}}=0[/latex].

Step 5. Draw the graph of [latex]y=A\csc(Bx)[/latex] but shift it up [latex]D=1[/latex].

Step 6. Sketch the vertical asymptotes, which occur at x = 0, x = 2, x = 4.

The graph for this function is shown in Figure 14.

A graph of 3 periods of a modified cosecant function, with 3 vertical asymptotes, and a dotted sinusoidal function that has local maximums where the cosecant function has local minimums and local minimums where the cosecant function has local maximums.

The vertical asymptotes shown on the graph mark off one period of the function, and the local extrema in this interval are shown by dots. Notice how the graph of the transformed cosecant relates to the graph of [latex]f(x)=2\sin\left(\frac{\pi}{2}x\right)+1[/latex], shown as the orange dashed wave.

Given the graph of [latex]f(x)=2\cos\left(\frac{\pi}{2}x\right)+1[/latex] shown in Figure 15, sketch the graph of [latex]g(x)=2\sec\left(\frac{\pi}{2}x\right)+1[/latex] on the same axes.

A graph of two periods of a modified cosine function. Range is [-1,3], graphed from x=-4 to x=4.

Analyzing the Graph of y = cot x and Its Variations

The last trigonometric function we need to explore is cotangent . The cotangent is defined by the reciprocal identity [latex]\cot x=\frac{1}{\tan x}[/latex]. Notice that the function is undefined when the tangent function is 0, leading to a vertical asymptote in the graph at 0, π, etc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers.

We can graph [latex]y=\cot x[/latex] by observing the graph of the tangent function because these two functions are reciprocals of one another. See Figure 16. Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of the tangent function increases, the graph of the cotangent function decreases.

The cotangent graph has vertical asymptotes at each value of x where [latex]\tan x=0[/latex]; we show these in the graph below with dashed lines. Since the cotangent is the reciprocal of the tangent, [latex]\cot x[/latex] has vertical asymptotes at all values of x where [latex]\tan x=0[/latex] , and [latex]\cot x=0[/latex] at all values of x where tan x has its vertical asymptotes.

A graph of cotangent of x, with vertical asymptotes at multiples of pi.

Figure 16. The cotangent function

A General Note: Features of the Graph of y = A cot( Bx )

  • The range is (−∞, ∞).
  • [latex]y=A\cot(Bx)[/latex] is an odd function.

Graphing Variations of y = cot x

We can transform the graph of the cotangent in much the same way as we did for the tangent. The equation becomes the following.

A General Note: Properties of the Graph of y = A cot( Bx −C)+ D

  • [latex]y=A\cot(Bx)[/latex] is an odd function because it is the quotient of even and odd functions (cosine and sine, respectively)

How To: Given a modified cotangent function of the form [latex]f(x)=A\cot(Bx)[/latex], graph one period.

  • Express the function in the form [latex]f(x)=A\cot(Bx)[/latex].
  • Identify the stretching factor, | A |.
  • Identify the period, [latex]P=\frac{\pi}{|B|}[/latex].
  • Draw the graph of [latex]y=A\tan(Bx)[/latex].
  • Plot any two reference points.
  • Use the reciprocal relationship between tangent and cotangent to draw the graph of [latex]y=A\cot(Bx)[/latex].

Example 10: Graphing Variations of the Cotangent Function

Determine the stretching factor, period, and phase shift of [latex]y=3\cot(4x)[/latex], and then sketch a graph.

Step 1. Expressing the function in the form [latex]f(x)=A\cot(Bx)[/latex] gives [latex]f(x)=3\cot(4x)[/latex].

Step 2. The stretching factor is [latex]|A|=3[/latex].

Step 3. The period is [latex]P=\frac{\pi}{4}[/latex].

Step 4. Sketch the graph of [latex]y=3\tan(4x)[/latex].

Step 5. Plot two reference points. Two such points are [latex]\left(\frac{\pi}{16}\text{, }3\right)[/latex] and [latex]\left(\frac{3\pi}{16}\text{, }−3\right)[/latex].

Step 6. Use the reciprocal relationship to draw [latex]y=3\cot(4x)[/latex].

Step 7. Sketch the asymptotes, [latex]x=0[/latex], [latex]x=\frac{\pi}{4}[/latex].

The orange graph in Figure 17 shows [latex]y=3\tan(4x)[/latex] and the blue graph shows [latex]y=3\cot(4x)[/latex].

A graph of two periods of a modified tangent function and a modified cotangent function. Vertical asymptotes at x=-pi/4 and pi/4.

How To: Given a modified cotangent function of the form [latex]f(x)=A\cot(Bx−C)+D[/latex], graph one period.

  • Express the function in the form [latex]f(x)=A\cot(Bx−C)+D[/latex].
  • Identify the phase shift, [latex]\frac{C}{B}[/latex].
  • Sketch the asymptotes [latex]x =\frac{C}{B}+\frac{\pi}{|B|}k[/latex], where k is an integer.

Example 11: Graphing a Modified Cotangent

Sketch a graph of one period of the function [latex]f(x)=4\cot\left(\frac{\pi}{8}x−\frac{\pi}{2}\right)−2[/latex].

Step 1. The function is already written in the general form [latex]f(x)=A\cot(Bx−C)+D[/latex].

Step 2.  [latex]A=4[/latex], so the stretching factor is 4.

Step 3.  [latex]B=\frac{\pi}{8}[/latex], so the period is [latex]P=\frac{\pi}{|B|}=\frac{\pi}{\frac{\pi}{8}}=8[/latex].

Step 4.  [latex]C=\frac{\pi}{2}[/latex], so the phase shift is [latex]\frac{C}{B}=\frac{\frac{\pi}{2}}{\frac{\pi}{8}}=4[/latex].

Step 5. We draw [latex]f(x)=4\tan\left(\frac{\pi}{8}x−\frac{\pi}{2}\right)−2[/latex].

Step 6-7. Three points we can use to guide the graph are (6,2), (8,−2), and (10,−6). We use the reciprocal relationship of tangent and cotangent to draw [latex]f(x)=4\cot\left(\frac{\pi}{8}x−\frac{\pi}{2}\right)−2[/latex].

Step 8. The vertical asymptotes are [latex]x=4[/latex] and [latex]x=12[/latex].

The graph is shown in Figure 18.

A graph of one period of a modified cotangent function. Vertical asymptotes at x=4 and x=12.

Figure 18. One period of a modified cotangent function.

Key Equations

Key concepts.

  • The tangent function has period π.
  • [latex]f(x)=A\tan(Bx−C)+D[/latex] is a tangent with vertical and/or horizontal stretch/compression and shift.
  • The secant and cosecant are both periodic functions with a period of2π. [latex]f(x)=A\sec(Bx−C)+D[/latex] gives a shifted, compressed, and/or stretched secant function graph.
  • [latex]f(x)=A\csc(Bx−C)+D[/latex] gives a shifted, compressed, and/or stretched cosecant function graph.
  • The cotangent function has period π and vertical asymptotes at 0, ±π,±2π,….
  • The range of cotangent is (−∞,∞),and the function is decreasing at each point in its range.
  • The cotangent is zero at [latex]\pm\frac{\pi}{2}\text{, }\pm\frac{3\pi}{2}[/latex],….
  • [latex]f(x)=A\cot(Bx−C)+D[/latex] is a cotangent with vertical and/or horizontal stretch/compression and shift.
  • Real-world scenarios can be solved using graphs of trigonometric functions.
  • Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Project : http://cnx.org/contents/[email protected]:1/Preface. License : CC BY: Attribution
  • Animation: Graphing the Tangent Function Using the Unit Circle. Authored by : Mathispower4u. Located at : https://youtu.be/ssjG9kE25OY . License : All Rights Reserved . License Terms : Standard YouTube License

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Trigonometric Functions

Trigonometric functions are the basic six functions that have a domain input value as an angle of a right triangle, and a numeric answer as the range. The trigonometric function (also called the 'trig function') of f(x) = sinθ has a domain, which is the angle θ given in degrees or radians, and a range of [-1, 1]. Similarly we have the domain and range from all other functions. Trigonometric functions are extensively used in calculus, geometry, algebra.

Here in the below content, we shall aim at understanding the trigonometric functions across the four quadrants, their graphs, the domain and range, the formulas, and the differentiation, integration of trigonometric functions. We will solve a few examples using these six trig functions for a better understanding of them and their applications.

What are Trigonometric Functions?

There are six basic trigonometric functions used in Trigonometry. These functions are trigonometric ratios. The six basic trigonometric functions are sine function , cosine function , secant function, co-secant function, tangent function , and co-tangent function. The trigonometric functions and identities are the ratio of sides of a right-angled triangle. The sides of a right triangle are the perpendicular side, hypotenuse, and base, which are used to calculate the sine, cosine, tangent, secant , cosecant , and cotangent values using trigonometric formulas.

Trigonometric Functions

Trigonometric Functions Formulas

We have certain formulas to find the values of the trig functions using the sides of a right-angled triangle. To write these formulas, we use the abbreviated form of these functions. Sine is written as sin, cosine is written as cos, tangent is denoted by tan, secant is denoted by sec, cosecant is abbreviated as cosec, and cotangent is abbreviated as cot. The basic formulas to find the trigonometric functions are as follows:

  • sin θ = Perpendicular/Hypotenuse
  • cos θ = Base/Hypotenuse
  • tan θ = Perpendicular/Base
  • sec θ = Hypotenuse/Base
  • cosec θ = Hypotenuse/Perpendicular
  • cot θ = Base/Perpendicular

As we can observe from the above-given formulas, sine and cosecant are reciprocals of each other. Similarly, the reciprocal pairs are cosine and secant, and tangent and cotangent.

Trigonometric Functions Values

The trigonometric functions have a domain θ, which is in degrees or radians . Some of the principal values of θ for the different trigonometric functions are presented below in a table . These principal values are also referred to as standard values of trig functions at specific angles and are frequently used in calculations. The principal values of trigonometric functions have been derived from a unit circle. These values also satisfy all the trigonometric formulas .

Trigonometric Functions Values

Trig Functions in Four Quadrants

The angle θ is an acute angle (θ < 90°) and is measured with reference to the positive x-axis, in the anticlockwise direction. Further, these trig functions have different numeric signs (+ or -) in the different quadrants, which are based on the positive or negative axis of the quadrant. The trigonometric functions of Sinθ, Cosecθ are positive in quadrants I and II, and are negative in quadrants III and IV. All the trigonometric functions have a positive range in the first quadrant. The trigonometric functions Tanθ, Cotθ are positive only in Quadrants I and III, and the trigonometric ratios of Cosθ, Secθ are positive only in quadrants I and IV.

Trigonometric Functions in Four Quadrants

The trigonometric functions have values of θ, (90° - θ) in the first quadrant. The cofunction identities provide the interrelationship between the different complementary trigonometric functions for the angle (90° - θ).

  • sin(90°−θ) = cos θ
  • cos(90°−θ) = sin θ
  • tan(90°−θ) = cot θ
  • cot(90°−θ) = tan θ
  • sec(90°−θ) = cosec θ
  • cosec(90°−θ) = sec θ

The domain θ value for different trigonometric function in the second quadrant is (π/2 + θ, π - θ), in the third quadrant is (π + θ, 3π/2 - θ), and in the fourth quadrant is (3π/2 + θ, 2π - θ). For π/2, 3π/2 the trigonometric values changes as their complementary ratios such as Sinθ⇔Cosθ, Tanθ⇔Cotθ, Secθ⇔Cosecθ. For π, 2π the trigonometric values remain the same. The changing trigonometric ratios in different quadrants and angles can be understood from the below table.

Trigonometric Functions Graph

The graphs of trigonometric functions have the domain value of θ represented on the horizontal x-axis and the range value represented along the vertical y-axis. The graphs of Sinθ and Tanθ pass through the origin and the graphs of other trigonometric functions do not pass through the origin. The range of Sinθ and Cosθ is limited to [-1, 1]. The range of infinite values is presented as drawn beside the dotted lines.

Graph of Trigonometric Functions

Domain and Range of Trigonometric Functions

The value of θ represents the domain of the trigonometric functions and the resultant value is the range of the trigonometric function. The domain values of θ are in degrees or radians and the range is a real number value. Generally, the domain of the trigonometric function is a real number value, but in certain cases, a few angle values are excluded because it results in a range as an infinite value. The trigonometric function are periodic functions . The below table presents the domain and range of the six trigonometric functions .

Trigonometric Functions Identities

The trigonometric functions identities are broadly divided into reciprocal identities, Pythagorean formulas, sum and difference of trig functions identities, formulas for multiple and sub-multiple angles, sum and product of identities. All of these below formulas can be easily derived using the ratio of sides of a right-angled triangle. The higher formulas can be derived by using the basic trigonometric function formulas. Reciprocal identities are used frequently to simplify trigonometric problems.

Reciprocal Identities

  • cosec θ = 1/sin θ
  • sec θ = 1/cos θ
  • cot θ = 1/tan θ
  • sin θ = 1/cosec θ
  • cos θ = 1/sec θ
  • tan θ = 1/cot θ

Pythagorean Identities

  • Sin 2 θ + Cos 2 θ = 1
  • 1 + Tan 2 θ = Sec 2 θ
  • 1 + Cot 2 θ = Cosec 2 θ

Sum and Difference Identities

  • sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
  • cos(x+y) = cos(x)cos(y) – sin(x)sin(y)
  • tan(x+y) = (tan x + tan y)/ (1−tan x tan y)
  • sin(x–y) = sin(x)cos(y) – cos(x)sin(y)
  • cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
  • tan(x−y) = (tan x–tan y)/ (1+tan x tan y)

Half-Angle Identities

  • sin A/2 = ±√[(1 - cos A) / 2]
  • cos A/2 = ±√[(1 + cos A) / 2]
  • tan A/2 = ±√[(1 - cos A) / (1 + cos A)] (or) sin A / (1 + cos A) (or) (1 - cos A) / sin A

Double Angle Identities

  • sin(2x) = 2sin(x) cos(x) = [2tan x/(1+tan 2 x)]
  • cos(2x) = cos 2 (x)–sin 2 (x) = [(1-tan 2 x)/(1+tan 2 x)]
  • cos(2x) = 2cos 2 (x)−1 = 1–2sin 2 (x)
  • tan(2x) = [2tan(x)]/ [1−tan 2 (x)]
  • cot(2x) = [cot 2 (x) - 1]/[2cot(x)]
  • sec (2x) = sec 2 x/(2-sec 2 x)
  • cosec (2x) = (sec x. cosec x)/2

Triple Angle Identities

  • Sin 3x = 3sin x – 4sin 3 x
  • Cos 3x = 4cos 3 x - 3cos x
  • Tan 3x = [3tanx-tan 3 x]/[1-3tan 2 x]

Product identities

  • 2sinx⋅cosy=sin(x+y)+sin(x−y)
  • 2cosx⋅cosy=cos(x+y)+cos(x−y)
  • 2sinx⋅siny=cos(x−y)−cos(x+y)

Sum of Identities

  • sinx+siny=2sin((x+y)/2) . cos((x−y)/2)
  • sinx−siny=2cos((x+y)/2) . sin((x−y)/2)
  • cosx+cosy=2cos((x+y)/2) . cos((x−y)/2)
  • cosx−cosy=−2sin((x+y)/2 . sin((x−y)/2)

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios . Here the basic trigonometric function of Sin θ = x, can be changed to Sin -1 x = θ. Here x can have values in whole numbers, decimals , fractions , or exponents . For θ = 30° we have θ = Sin -1 (1/2). All the trigonometric formulas can be transformed into inverse trigonometric function formulas.

Arbitrary Values: The inverse trigonometric ratio formula for arbitrary values is applicable for all the six trigonometric functions. For the inverse trigonometric functions of sine, tangent, cosecant, the negative of the values are translated as the negatives of the function. And for functions of cosecant, secant, cotangent, the negatives of the domain are translated as the subtraction of the function from the π value.

  • Sin -1 (-x) = -Sin -1 x
  • Tan -1 (-x) = -Tan -1 x
  • Cosec -1 (-x) = -Cosec -1 x
  • Cos -1 (-x) = π - Cos -1 x
  • Sec -1 (-x) = π - Sec -1 x
  • Cot -1 (-x) = π - Cot -1 x

The inverse trigonometric functions of reciprocal and complementary functions are similar to the basic trigonometric functions. The reciprocal relationship of the basic trigonometric functions, sine-cosecant, cos-secant, tangent-cotangent, can be interpreted for the inverse trigonometric functions. Also the complementary functions, since-cosine, tangent-cotangent, and secant-cosecant can be interpreted into:

Reciprocal Functions: The inverse trigonometric formula of inverse sine, inverse cosine, and inverse tangent can also be expressed in the following forms.

  • Sin -1 x = Cosec -1 1/x
  • Cos -1 x = Sec -1 1/x
  • Tan -1 x = Cot -1 1/x

Complementary Functions: The complementary functions of sine-cosine, tangent-cotangent, secant-cosecant, sum up to π/2.

  • Sin -1 x + Cos -1 x = π/2
  • Tan -1 x + Cot -1 x = π/2
  • Sec -1 x + Cosec -1 x = π/2

Trigonometric Functions Derivatives

The differentiation of trigonometric functions gives the slope of the tangent of the curve. The differentiation of Sinx is Cosx and here on applying the x value in degrees for Cosx we can obtain the slope of the tangent of the curve of Sinx at a particular point. The formulas of differentiation of trigonometric functions are useful to find the equation of a tangent, normal, to find the errors in calculations.

  • d/dx. Sinx = Cosx
  • d/dx. Cosx = -Sinx
  • d/dx. Tanx = Sec 2 x
  • d/dx. Cotx = -Cosec 2 x
  • d/dx.Secx = Secx.Tanx
  • d/dx. Cosecx = - Cosecx.Cotx

Integration of Trigonometric Function

The integration of trigonometric functions is helpful to find the area under the graph of the trigonometric function. Generally, the area under the graph of the trigonometric function can be calculated with reference to any of the axis lines and within a defined limit value. The integration of trigonometric functions is helpful to generally find the area of irregularly shaped plane surfaces.

  • ∫ cosx dx = sinx + C
  • ∫ sinx dx = -cosx + C
  • ∫ sec 2 x dx = tanx + C
  • ∫ cosec 2 x dx = -cotx + C
  • ∫ secx.tanx dx = secx + C
  • ∫ cosecx.cotx dx = -cosecx + C
  • ∫ tanx dx = log|secx| + C
  • ∫ cotx.dx = log|sinx| + C
  • ∫ secx dx = log|secx + tanx| + C
  • ∫ cosecx.dx = log|cosecx - cotx| + C

Related Topics

The following related links help in understanding more about trigonometric identities.

  • Trigonometry
  • Sum to Product Formulas
  • Algebraic Identities

Solved Examples on Trigonometric Functions

Example 1: Find the value of Sin75°.

The aim is to find the value of Sin75°.

Her we can use the formula Sin(A + B) = SinA.CosB + CosA.SinB.

Here we have A = 30° and B = 45°

Sin 75° = Sin(30° + 45°)

= Sin30°.Cos45° + Cos30°.Sin45°

= (1/2) (1/√2) + (√3/2) (1/√2)

= 1/2√2 + √3/2√2

= (√3 + 1) / 2√2

Answer: Sin75° = (√3 + 1) / 2√2

Example 2: Find the value of the trigonometric functions, for the given value of 12Tanθ = 5.

Given 12Tanθ = 5, and we have Tanθ = 5/12

Tanθ = Perpendicular/Base = 5/12

Applying the Pythagorean theorem we have:

Hypotenuse 2 = Perpendicular 2 + Base 2

Hyp 2 = 12 2 + 5 2

Hence the other trigonometric functions are as follows.

Sinθ = Perp/Hyp = 5/13

Cosθ = Base/Hyp = 12/13

Cotθ = Base/Perp = 12/5

Secθ = Hyp/Base = 13/12

Cosecθ = Hyp/Perp = 13/5

Example 3: Find the value of the product of the six trigonometric functions.

Solution: We know that cosec x is the reciprocal of sin x and sec x is the reciprocal of cos x. Also, tan x can be written as the ratio of sin x and cos x, cot x can be written as the ratio of cos x and sin x. So, we have

sinx × cosx × tanx × cotx × secx × cosecx = sinx × cosx × (sinx/cosx) × (cosx/sinx) × (1/cosx) × (1/sinx)

= (sinx × cosx) / (sinx × cosx) × (sinx/cosx) × (cosx/sinx)

Answer: Product of the six trigonometric functions is equal to 1.

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graphing other trigonometric functions solving real world problems

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Practice Questions on Trigonometric Functions

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FAQs on Trigonometric Functions

What are the six trigonometric functions.

The trigonometric functions are the result of the ratio of the sides of the right angles triangle. For the three sides of the triangle as hypotenuse, base, altitude, and for the angle between the hypotenuse and the base being θ, the value of the six trigonometric ratios is as follows.

  • Sinθ = Altitude/Hypotenuse
  • Cosθ = Base/Hypotenuse
  • Tanθ = Altitude/Base
  • Cotθ = Base/Altitude
  • Secθ = Hypotenuse/Base
  • Cosecθ = Hypotenuse/Altitude

How Do you Find Trigonometric Functions?

The trigonometric functions are the ratio of the sides of a right-angled triangle . Further, we also apply the Pythagorean rule of Hypotenuse 2 = Altitude 2 + Base 2 . Also, the trigonometric functions have different values for different angle values between the hypotenuse and the base of the right triangle.

What is the Domain and Range of Trigonometric Functions?

The domain of a trigonometric function is the value of θ in Sinθ, and the range is the final numeric value of Sinθ. This concept can be similarly applied to all the other trigonometric functions. Further, the domain values can be any angular values, but here we have the principal values of angles as 0°, 30°, 45°, 60°, 90°. And the range is the highest and the lowest values which are obtained. It is [-1, 1] for sinθ, cosθ, and it is (-∞, +∞) for tanθ, cotθ.

What is the Result of Multiplying Six Trigonometric Functions?

The result of the multiplication of the six trigonometric functions is as follows. Sinθ.Cosθ.Tanθ.Cotθ.Secθ.Cosecθ = Sinθ.Cosθ.Sinθ/Cosθ.Cosθ/Sinθ.1/Cosθ.1/Sinθ = 1.

What is the General Solution of Trigonometric Function of Sinx?

The general solution of Sinx is nπ + (-1) n x. This represents all the higher angle values of Sinx. For x = π/3 we have the higher values of x as 2π/3, 7π3, and the general solution of x is nπ +(-1) n π/3.

What is the General Solution of the Trigonometric Function of Cosx?

The general solution of Cosx is 2nπ + x. This general solution represents all the higher angle values of Cosx. For x = π/4, the higher values of x are 7π/4, 9π/4, and the general solution of x is 2nπ + π/4.

What is the General Solution of the Trig Function of Tanx?

The general solution of Tanx is nπ + x. The general solution represents all the higher angle values of Tanx. For x = π/6, the higher values of x are 7π/6, 13π/6, and the general solution of x is nπ + π/6.

How to Differentiate Trigonometric Functions?

The differentiation of trigonometric function results in the slope of the tangent to the curve of the trigonometric function. The differentiation of sinx results in cosx, which by substituting the value of x in degrees gives the slope value of the tangent to the curve of sinx. The differentiation is calculated using the first principle of derivatives. Further, we have the differentiation of the six trigonometric functions as follows.

What are the Applications of Trigonometric Functions?

The trigonometric functions have numerous applications in calculus coordinate geometry algebra. The slope of a line , the normal form of the equation of a lie, parametric coordinates of a parabola , ellipse , hyperbola, are all calculated and represented using trigonometric functions. The trigonometric functions can be used to find the height of a tree, for the given distance of the tree from the point of observation. Further, the trigonometric functions are extensively used in astronomy, to find distances of stars and celestial bodies, with the help of the given angle value.

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Chapter 4: Trig Functions

Exercises: 4.2 Graphs of Trigonometric Functions

                         skills.

Practice each skill in the Homework Problems listed:

  • Find coordinates
  • Use bearings to determine position
  • Sketch graphs of the sine and cosine functions
  • Find the coordinates of points on a sine or cosine graph
  • Use function notation
  • Find reference angles
  • Solve equations graphically
  • Graph the tangent function
  • Find and use the angle of inclination of a line

Suggested Problems

Homework 4.2

Exercise group..

For Problems 1–6, find exact values for the coordinates of the point.

For Problems 7–12, find the coordinates of the point, rounded to hundredths.

For Problems 13–18, a ship sails from the seaport on the given bearing for the given distance.

  • Make a sketch showing the ship’s current location relative to the seaport.
  • How far east or west of the seaport is the ship’s present location? How far north or south?

[latex]36°{,}[/latex] 26 miles

[latex]124°{,}[/latex] 80 km

[latex]230°{,}[/latex] 120 km

[latex]318°{,}[/latex] 75 miles

[latex]285°{,}[/latex] 32 km

[latex]192°{,}[/latex] 260 miles

  • Draw vertical line segments from the unit circle to the [latex]x[/latex]-axis to illustrate the [latex]y[/latex]-coordinate of each point designated by the angles, [latex]0°[/latex] to [latex]90°{,}[/latex] shown on the figure below.
  • Transfer your vertical line segments to the appropriate position on the grid below.
  • Repeat parts (a) and (b) for the other three quadrants.
  • Connect the tops of the segments to sketch a graph of [latex]y = \sin \theta{.}[/latex]
  • Draw horizontal line segments from the unit circle to the [latex]y[/latex]-axis to illustrate the [latex]x[/latex]-coordinate of each point designated by the angles, [latex]0°[/latex] to [latex]90°{,}[/latex] shown on the figure below.
  • Transfer your horizontal line segments into vertical line segments at the appropriate position on the grid below.
  • Connect the tops of the segments to sketch a graph of [latex]y = \cos \theta{.}[/latex]
  • Prepare a graph with the horizontal axis scaled from [latex]0°[/latex] to [latex]360°[/latex] in multiples of [latex]45°{.}[/latex]
  • Sketch a graph of [latex]f(\theta) = \sin \theta[/latex] by plotting points for multiples of [latex]45°{.}[/latex]
  • Sketch a graph of [latex]f(\theta) = \cos \theta[/latex] by plotting points for multiples of [latex]45°{.}[/latex]
  • Prepare a graph with the horizontal axis scaled from [latex]0°[/latex] to [latex]360°[/latex] in multiples of [latex]30°{.}[/latex]
  • Sketch a graph of [latex]f(\theta) = \cos \theta[/latex] by plotting points for multiples of [latex]30°{.}[/latex]
  • Sketch a graph of [latex]f(\theta) = \sin \theta[/latex] by plotting points for multiples of [latex]30°{.}[/latex]

For Problems 27–30, give the coordinates of each point on the graph of [latex]f(\theta) = \sin \theta[/latex] or [latex]f(\theta) = \cos \theta[/latex]-27.

Make a short table of values like the one shown, and sketch the function by hand. Be sure to label the [latex]x[/latex]-axis and [latex]y[/latex]-axis appropriately.

  • [latex]\displaystyle f(\theta) = \sin \theta[/latex]
  • [latex]\displaystyle f(\theta) = \cos \theta[/latex]

For Problems 33–40, evaluate the expression for [latex]f(\theta) = \sin \theta[/latex] and [latex]g(\theta) = \cos \theta{.}[/latex]

[latex]3 + f(30°)[/latex]

[latex]3 f(30°)[/latex]

[latex]4g(225°) - 1[/latex]

[latex]-4 + 2g(225°) - 1[/latex]

[latex]-2f(3\theta){,}[/latex] for [latex]\theta = 90°[/latex]

[latex]6f(\dfrac{\theta}{2}){,}[/latex] for [latex]\theta = 90°[/latex]

[latex]8 - 5g(\dfrac{\theta}{3}){,}[/latex] for [latex]\theta = 360°[/latex]

[latex]1 - 4g(4\theta){,}[/latex] for [latex]\theta = 135°[/latex]

Draw two different angles [latex]\alpha[/latex] and [latex]\beta[/latex] in standard position whose sine is [latex]0.6{.}[/latex]

  • Use a protractor to measure [latex]\alpha[/latex] and [latex]\beta{.}[/latex]
  • Find the reference angles for both [latex]\alpha[/latex] and [latex]\beta{.}[/latex] Draw in the reference triangles.

Draw two different angles [latex]\theta[/latex] and [latex]\phi[/latex] in standard position whose sine is [latex]-0.8{.}[/latex]

  • Use a protractor to measure [latex]\theta[/latex] and [latex]\phi{.}[/latex]
  • Find the reference angles for both [latex]\theta[/latex] and [latex]\phi{.}[/latex] Draw in the reference triangles.

Draw two different angles [latex]\alpha[/latex] and [latex]\beta[/latex] in standard position whose cosine is [latex]0.3{.}[/latex]

Draw two different angles [latex]\theta[/latex] and [latex]\phi[/latex] in standard position whose cosine is [latex]-0.4{.}[/latex]

[latex]\sin \theta = 0.6[/latex]

[latex]\sin \theta = -0.8[/latex]

[latex]\cos \theta = 0.3[/latex]

[latex]\cos \theta = -0.4[/latex]

[latex]\sin \theta = -0.2[/latex]

[latex]\sin \theta = 1.2[/latex]

[latex]\cos \theta = -0.9[/latex]

[latex]\cos \theta = -1.1[/latex]

  • What happens to [latex]\tan \theta[/latex] as [latex]\theta[/latex] increases toward [latex]90°{?}[/latex]
  • What happens to [latex]\tan \theta[/latex] as [latex]\theta[/latex] decreases toward [latex]90°{?}[/latex]
  • What value does your calculator give for [latex]\tan 90°{?}[/latex] Why?
  • Sketch by hand a graph of [latex]y = \tan \theta[/latex] for [latex]-180° \le \theta \le 180°{.}[/latex]
  • Use your calculator to graph [latex]y = \tan \theta[/latex] in the ZTrig window (press ZOOM 7). Sketch the result. On your sketch, mark scales on the axes and include dotted lines for the vertical asymptotes.

For Problems 61–64,find the angle of inclination of the line.

[latex]y = \dfrac{5}{4}x - 3[/latex]

[latex]y = 6 + \dfrac{2}{9}x[/latex]

[latex]y = -2 - \dfrac{3}{8}x[/latex]

[latex]y = \dfrac{-7}{2}x + 1[/latex]

For Problems 65–68, find an equation for the line passing through the given point with angle of inclination [latex]\alpha{.}[/latex]

[latex](3,-5), ~\alpha = 28°[/latex]

[latex](-2,6), ~\alpha = 67°[/latex]

[latex](-8,12), ~\alpha = 112°[/latex]

[latex](-4,-1), ~\alpha = 154°[/latex]

The slope of a line is a function of its angle of inclination, [latex]m = f(\alpha){.}[/latex] Complete the table and sketch a graph of the function.

  • What happens to the slope of the line as [latex]\alpha[/latex] increases toward [latex]90°{?}[/latex]
  • What happens to the slope of the line as [latex]\alpha[/latex] decreases toward [latex]90°{?}[/latex]

The angle of inclination of a line is a function of its slope, [latex]\alpha = g(m){.}[/latex] Complete the table and sketch a graph of the function.

  • What happens to the angle of inclination as the slope increases toward infinity?
  • What happens to the angle of inclination as the slope decreases toward negative infinity?

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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Mathematics LibreTexts

2.6: Solving Trigonmetric Equations

  • Last updated
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  • Page ID 7109

  • Ted Sundstrom & Steven Schlicker
  • Grand Valley State University via ScholarWorks @Grand Valley State University

Focus Questions

The following questions are meant to guide our study of the material in this section. After studying this section, we should understand the concepts motivated by these questions and be able to write precise, coherent answers to these questions.

For these questions, we let \(q\) be a real number with \(-1 \leq q \leq 1\) and let \(r\) be a real number.

  • How can an inverse trigonometric function be used to determine one solution of an equation of the form \(\sin(x) = q, \cos(x) = q\), or \(\tan(x) = r\)?
  • How can properties of the trigonometric functions be used to determine all solutions of an equation of the form \(\sin(x) = q, \cos(x) = q\), or \(\tan(x) = r\)? within one complete period of the trigonometric function?
  • How can we use the period of a trigonometric function to determine a formula for the solutions of an equation of the form \(\sin(x) = q, \cos(x) = q\), or \(\tan(x) = r\)?

Recall that a mathematical equation like \(x^{2} = 1\) is a relation between two expressions that may be true for some values of the variable while an identity like \(\cos(-x) = \cos(x)\) is an equation that is true for all allowable values of the variable. So an identity is a special type of equation. Equations that are not identities are also called conditional equations because they are not valid for all allowable values of the variable. To solve an equation means to find all of the values for the variables that make the two expressions on either side of the equation equal to each other. We solved algebraic equations in algebra and now we will solve trigonometric equations.

A trigonometric equation is an equation that involves trigonometric functions. We have already used graphical methods to approximate solutions of trigonometric equations. In Example 2.17, we used the function

\[V(t) = 35\cos \left(\dfrac{5\pi}{3}t \right) + 105\]

as a model for the amount of blood in the heart. For this function, t is measured in seconds since the heart was full and V .t / is measured in milliliters. To determine the times when there are 140 milliliters of blood in the heart, we needed to solve the equation

\[35\cos \left(\dfrac{5\pi}{3}t \right) + 105 = 100\]

At that time, we used the “intersect” capability of a graphing utility to determine some solutions of this equation. In this section, we will learn how to use the inverse cosine function and properties of the cosine function to determine the solutions of this equation. We begin by first studying simpler equations.

Beginning Activity

Use a graphing utility to draw the graphs of \(y = \cos(x)\) and \(y = 0.7\) on the same axes using \(-\pi \leq x \leq \pi\) and \(-1.2 \leq y \leq 1.2\). Use the graphing utility to find the points of intersection of these two graphs and to determine solutions of the equation \(\cos(x) = 0.7\)

In the beginning activity, we should have determined the following approximations for solutions of the equation \(\cos(x) = 0.7\):

  • \(x_{1} \approx 0.79540\)
  • \(x_{1} \approx -0.79540\)

These approximations have been rounded to five decimal places.

The graph below shows the two graphs using \(-3\pi \leq x \leq 3\pi\). The solutions \(x_{1}\) and \(x_{2}\) are shown on the graph. As can be seen, the graph shows \(x_{1}\) and \(x_{2}\) and four other solutions to the equation \(\cos(x) = 0.7\). In fact, if we imagine the graph extended indefinitely to the left and to the right, we can see that there are infinitely many solutions for this equation.

154898249770254270.png

This is where we can use the fact that the period of the cosine function is \(2\pi\). The other solutions differ from \(x_{1}\) or \(x_{2}\) by an integer multiple of the period of \(2\pi\). We can represent an integer multiple of \(2\pi\) by \(k(2\pi)\) for some integer \(k\). So we say that any solution of the equation \(\cos(x) = 0.7\) can be approximated by

\[x_{1} \approx 0.79540 + k(2\pi)\]

\[x_{1} \approx -0.79540 + k(2\pi)\]

For example, if we use \(k = 4\), we see that \(x \approx 25.92814\) or \(x \approx 24.33734\).

We can use a calculator to check that for both values, \(\cos(x) = 0.7\).

A Strategy for Solving a Trigonometric Equation

The example using the equation \(\cos(x) = 0.7\) was designed to illustrate the fact that if there are no restrictions placed on the unknown \(x\), then there can be infinitely many solutions for an equation of the form “some trigonometric function of \(x\)” = a number.

A general strategy to solve such equations is:

  • Find all solutions of the equation within one period of the function. This is often done by using properties of the trigonometric function. Quite often, there will be two solutions within a single period.
  • Use the period of the function to express formulas for all solutions by adding integer multiples of the period to each solution found in the first step. For example, if the function has a period of \(2\pi\) and \(x_{1}\) and \(x_{2}\) are the only two solutions in a complete period, then we would write the solutions for the equation as \[x = x_{1} + k(2\pi), x = x_{2} + k(2\pi)\], where \(k\) is an integer.

Instead of writing “\(k\) is an integer,” we could write \[k \in \{\dotsb, -2, -1, 0, 1, 2, \dotsb\}.\]

Exercise \(\PageIndex{1}\)

Use a graph to approximate the solutions (rounded to four decimal places) of the equation \(\sin(x) = -0.6\) on the interval \(-\pi \leq x \leq \pi\). Then use the period of the sine function to write formulas that can be used to approximate any solution of this equation.

Any solution of the equation \(\sin(x) = -0.6\) may be approximated with one of the following:

\[x \approx 0.64350 + k(2\pi)\] or \[x \approx 2.49809 + k(2\pi)\]

Using Inverse Functions to Solve Trigonometric Equations

Although we can use a graphing utility to determine approximations for solutions to many equations, we often need to have some notation to indicate specific numbers (that are often solutions of equations). We have already seen this in previous mathematics courses. For example, we use the notation \(\sqrt{20}\) to represent the positive real number whose square is equal to \(20\). We can use this to say that the two solutions of the equation \(x^{2} = 20\) are \[x = \sqrt{20} \space and \space x = -\sqrt{20}\]

Notice that there are two solutions of the equation but \(\sqrt{20}\) represents only one of those solutions. We will now learn how to use the inverse trigonometric functions to do something similar for trigonometric equations. One big difference is that most trigonometric equations will have infinitely many solutions instead of just two. We will use the inverse trigonometric functions to represent one solution of an equation and then learn how to represent all of the solutions in terms of this one solution. We will first show how this is done with the equation \(\cos(x) = 0.7\) from the beginning activity for this section.

Example \(\PageIndex{1}\): Solving an Equation Involving the Cosine Function

For the equation \(\cos(x) = 0.7\), we first use the result about the inverse cosine function on page 150, which states that for \(t\) in the closed interval \([0, \pi]\), \[\cos^{-1}(\cos(t)) = t\]

So we “apply the inverse cosine function” to both sides of the equation \(\cos(x) = 0.7\) This gives:

\[\cos(x) = 0.7\]

\[\cos^{-1}(\cos(x)) = \cos^{-1}(0.7)\]

\[x = \cos^{-1}(0.7) \nonumber\]

Another thing we must remember is that this gives the one solution for the equation that is in interval \([0, \pi]\). Before we use the periodic property, we need to determine the other solutions for the equation in one complete period of the cosine function. We can use the interval \([0, 2\pi]\) but it is easier to use the interval \([-\pi, \pi]\). One reason for this is the following so-called “negative arc identity” stated on p age 82.

\[\cos(-x) = \cos(x)\) for every real number \(x\).

Hence, since one solution for the equation is \(x = \cos^{-1}(0.7)\), another solution is \(x = -\cos^{-1}(0.7)\). This means that the two solutions of the equation \(x = \cos(x)\) on the interval \([-\pi, \pi]\) are

\[x = \cos^{-1}(0.7) \space and \space x = -\cos^{-1}(0.7) \nonumber\]

It can be verified that the equation \(\cos(x) = 0.7\) has two solutions on the interval \([-\pi, \pi]\) by drawing the graphs of \(y = \cos(x)\) and \(y = 0.7\) on the interval \([-\pi, \pi]\). So if we restrict ourselves to this interval, we have something very much like solving the equation \(x^{2} = 20\) in that there are two solutions that are negatives of each other. The main difference now is that the trigonometric equation has infinitely many solutions and as before, we now use the periodic property of the cosine function. Since the period is \(2\pi\), just like with the numerical approximations from the beginning activity, we can say that any solution of the equation \(\cos(x) = 0.7\) will be of the form

\[x = \cos^{-1}(0.7) + k(2\pi) \space or \space x = -\cos^{-1}(0.7) + k(2\pi) \nonumber\]

where \(k\) is some integer.

Exercise \(\PageIndex{2}\)

Determine all solutions of the equation \(4\cos(x) + 3 = 2\) in the interval \([-\pi, \pi]\). Then use the periodic property of the cosine function to write formulas that can be used to generate all the solutions of this equation.

First use algebra to rewrite the equation in the form \(\cos(x)\) = “some number”.

We first rewrite the equation \(4\cos(x) + 3 = 2\) as follows:

\[\begin{align*} 4\cos(x) + 3 &= 2 \\[4pt] 4\cos(x) &= -1 \\[4pt] \cos(x) &= -\dfrac{1}{4} \end{align*}\]

So in the interval \([-\pi, \pi]\), the solutions are \(x_{1} = \arccos(-\dfrac{1}{4})\) and \(x_{2} = -\arccos(-\dfrac{1}{4})\). So any solution of the equation \(4\cos(x) + 3 = 2\) is of the form

\[x = \arccos(-\dfrac{1}{4}) + k(2\pi)\] or \[x = -\arccos(-\dfrac{1}{4}) + k(2\pi) \nonumber\]

The previous examples have shown that when using the inverse cosine function to solve equations of the form \(\cos(x)\) = a number, it is easier to use the interval \([-\pi, \pi]\); rather than the interval \([0, 2\pi]\). This is not necessarily true when using the inverse sine function since the inverse sine function gives a value in the interval \([-\dfrac{\pi}{2}, \dfrac{\pi}{2}]\). However, to keep things similar, we will continue to use the interval \([-\pi, \pi]\); as the complete period for the sine (or cosine) function. For the inverse sine, we use the following property stated on page 147.

For each t in the closed interval \([-\dfrac{\pi}{2}, \dfrac{\pi}{2}]\), \[\sin^{-1}(sin(t)) = t\]

When solving equations involving the cosine function, we also used a negative arc identity. We do the same and will use the following negative arc identity stated on page 82. \[\sin(-x) = -\sin(x)\] for every real number \(x\).

Example \(\PageIndex{2}\): Solving an Equation Involving the Sine Function

We will illustrate the general process using the equation \(\sin(x) = -0.6\) from Progress Check 2.34. Because of the negative arc identity for the sine function, it is actually easier to work with the equation \(\sin(x) = 0.6\). This is because if \(x = a\) is a solution of the equation \(\sin(x) = 0.6\), then \[\sin(-a) = -\sin(a) = -0.6\]

and so, \(x = -a\) is a solution of the equation \(\sin(x) = -0.6\). For the equation \(\sin(x) = 0.6\), we start by “applying the inverse sine function” to both sides of the equation.

\[\sin(x) = 0.6\] \[\sin^{-1}(\sin(x)) = \sin^{-1}(0.6)\] \[x = \sin^{-1}(0.6)\]

We need to remember that this is only one solution of the equation. Since we know that the sine function is positive in the first and second quadrants, this solution is in the first quadrant and there is another solution in the second quadrant. Using \(x = \sin^{-1}(0.6)\) as a reference arc (angle), the solution in the second quadrant is\(x = \pi - \sin^{-1}(0.6)\). We now use the result that if \(x = a\) is a solution of the equation \(\sin(x) = 0.6\), then \(x = -a\) is a solution of the equation \(\sin(x) = -0.6\). Please note that \[-(\pi - \sin^{-1}(0.6)) = -\pi + \sin^{-1}(0.6)\]

Our work so far is summarized in the following table.

At this point, we should use a calculator to verify that the two values in the right column are actually solutions of the equation \(\sin(x) = -0.6\). Now that we have the solutions for \(\sin(x) = -0.6\) in one complete cycle, we can use the fact that the period of the sine function is \(2\pi\) and say that the solutions of the equation \(\sin(x) = -0.6\) have the form

\[x = -\sin^{-1}(0.6) + k(2\pi) \space or \space x = (-\pi + \sin^{-1}(0.6)) + k(2\pi) \nonumber\]

Exercise \(\PageIndex{3}\)

Determine all solutions of the equation \(2\sin(x) + 1.2 = 2.5\) in the interval \([-\pi, \pi]\) Then use the periodic property of the sine function to write formulas that can be used to generate all the solutions of this equation.

Hint : First use algebra to rewrite the equation in the form \(sin(x)\) = “some number”.

We first use algebra to rewrite the equation \(2\sin(x) + 1.2 = 2.5\) in the form

\[\sin(x) = 0.65\]

So in the interval \([-\pi, \pi]\), the solutions are \(x_{1} = \arcsin(0.65)\) and \(x_{2} = \pi - \arcsin(0.65)\). So any solution of the equation \(2\sin(x) + 1.2 = 2.5\) is of the form

\[x = \arcsin(0.65) + k(2\pi)\] or \[x = \pi - \arcsin(0.65) + k(2\pi)\]

Solving More Complicated Trigonometric Equations

We have now learned to solve equations of the form \(\cos(x) = q\), and \(\sin(x) = q\), where \(q\) is a real number and \(-1 \leq q \leq 1\). We can use our ability to solve these types of equations to help solve more complicated equations of the form \(\cos(f(x)) = q\), and \(\sin(f(x)) = q\) where \(f\) is some function. The idea (which is typical in mathematics) is to convert this more complicated problem to two simpler problems. The idea is to:

  • Make the substitution \(t = f(x)\) to get an equation of the form \(\cos(t) = q\), or \(\sin(t) = q\).
  • Solve the equation in (1) for \(t\).
  • For each solution \(t\) of the equation in (1), solve the equation \(f(x) = t\) for \(x\). This step may be easy, difficult, or perhaps impossible depending on the equation \(f(x) = t\).

This process will be illustrated in the next progress check, which will be a guided investigation for solving the equation \(3\cos(2x + 1) + 6 = 5\).

Exercise \(\PageIndex{4}\)

We will solve the equation \(3\cos(2x + 1) + 6 = 5\).

  • First, use algebra to rewrite the equation in the form \(\cos(2x + 1) = -\dfrac{1}{3}\). Then, make the substitution \(t = 2x + 1\).
  • Determine all solutions of the equation \(\cos(t) = -\dfrac{1}{3}\) with \(-\pi \leq t \leq \pi\).
  • For each of these two solutions, use \(t = 2x + 1\) to find corresponding solutions for \(x\). In addition, use the substitution \(t = 2x + 1\) to write \(-\pi \leq 2x + 1 \leq \pi\) and solve this inequality for \(x\). This will give all of the solutions of the equation \(\cos(2x + 1) = -\dfrac{1}{3}\) in one complete cycle of the function given by \(y = \cos(2x + 1)\)
  • What is the period of the function \(y = \cos(2x + 1)\). Use the results in (3) and this period to write formulas that will generate all of the solutions of the equation \(\cos(2x + 1) = -\dfrac{1}{3}\). These will be the solutions of the original equation \(3\cos(2x + 1) + 6 = 5\).

1. \[3\cos(2x + 1) + 6 = 5\] \[3\cos(2x + 1) = -1\] \[\cos(2x + 1) = -\dfrac{1}{3}\]

2. \(t = \cos^{-1}(-\dfrac{1}{3})\) or \(t = -\cos^{-1}(-\dfrac{1}{3})\).

3. \[2x + 1 = \cos^{-1}(-\dfrac{1}{3})\]

\[2x = \cos^{-1}(-\dfrac{1}{3}) - 1\]

\[x = \dfrac{1}{2}\cos^{-1}(-\dfrac{1}{3}) - \dfrac{1}{2}\]

\[2x + 1 = -\cos^{-1}(-\dfrac{1}{3})\]

\[2x = -\cos^{-1}(-\dfrac{1}{3}) - 1\]

\[x = -\dfrac{1}{2}\cos^{-1}(-\dfrac{1}{3}) - \dfrac{1}{2}\]

4. The period of the function \(y = \cos(2x + 1)\). So the following formulas can be used to generate the solutions for the equation.

\[x = (\dfrac{1}{2}\cos^{-1}(-\dfrac{1}{3}) - \dfrac{1}{2}) + k\pi\] or \[x = (-\dfrac{1}{2}\cos^{-1}(-\dfrac{1}{3}) - \dfrac{1}{2}) + k\pi\]

where \(k\) is some integer. Notice that we added an integer multiple of the period, which is \(\pi\), to the solutions in (3).

Solving Equations Involving the Tangent Function

Solving an equation of the form \(\tan(x) = q\) is very similar to solving equations of the form \(\cos(x) = q\) or \(\sin(x) = q\). The main differences are the tangent function has a period of \(\pi\) (instead of \(2\pi\)), and the equation \(\tan(x) = q\) has only one solution in a complete period. We, of course, use the inverse tangent function for the equation \(\tan(x) = q\).

Exercise \(\PageIndex{5}\)

Use the inverse tangent function to determine one solution of the equation \[4\tan(x) + 1 = 10\] in the interval \((-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2})\). Then determine a formula that can be used to generate all solutions of this equation.

We first write the equation \(4\tan(x) + 1 = 10\) in the form \(\tan(x) = \dfrac{9}{4}\). So the only solution of the equation in the interval \(-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}\) is

\[x = \arctan(\dfrac{9}{4})\]

Since the period of the tangent function is \(\pi\), any solution of this equation can be written in the form \[x = \arctan(\dfrac{9}{4}) + k\pi\] where \(k\) is some integer.

A trigonometric equation is an equation that involves trigonometric functions. If we can write the trigonometric equation in the form “some trigonometric function of \(x\)” = a number, then we can use the following strategy to solve the equation.

  • Find one solution of the equation using the appropriate inverse trigonometric function.
  • Determine all solutions of the equation within one complete period of the trigonometric function. (This often involves the use of a reference arc based on the solution obtained in the first step.)
  • Use the period of the function to write formulas for all of the solutions of the trigonometric equation.

IMAGES

  1. Solving Trigonometric Equations that Model a Real-world Situation

    graphing other trigonometric functions solving real world problems

  2. Solving problems using trigonometric functions

    graphing other trigonometric functions solving real world problems

  3. Graphing Trigonometric Functions

    graphing other trigonometric functions solving real world problems

  4. Solving Trigonometric Equations Using Identities and Substitution

    graphing other trigonometric functions solving real world problems

  5. Problem Solving With Trigonometric Functions

    graphing other trigonometric functions solving real world problems

  6. Solving Trigonometric Equations By Finding All Solutions

    graphing other trigonometric functions solving real world problems

VIDEO

  1. Graphs of Other Trigonometric Functions

  2. Using Trigonometric Functions

  3. Trick for graphing Trigonometric Functions #viral

  4. Algebra 2 10.5 Graphing Other Trigonometric Functions

  5. PreCalculus II (Trig)

  6. 1st Secondary

COMMENTS

  1. 8.2: Graphs of the Other Trigonometric Functions

    Real-world scenarios can be solved using graphs of trigonometric functions. See Example \(\PageIndex{10}\). This page titled 8.2: Graphs of the Other Trigonometric Functions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts ...

  2. 6.2 Graphs of the Other Trigonometric Functions

    Analyzing the Graphs of y = sec x and y = cscx. The secant was defined by the reciprocal identity sec x = 1 cos x. sec x = 1 cos x. Notice that the function is undefined when the cosine is 0, leading to vertical asymptotes at π 2, π 2, 3 π 2, 3 π 2, etc. Because the cosine is never more than 1 in absolute value, the secant, being the reciprocal, will never be less than 1 in absolute value.

  3. Section 4.6: Graphs of the Other Trigonometric Functions

    Determine a function formula from a tangent or cotangent graph. Analyze the graphs of y=sec x and y=csc x. Graph variations of y=sec x and y=csc x. Determine a function formula from a secant or cosecant graph. Analyzing the Graph of y = tan x and Its Variations

  4. 8.3: Graphs of the Other Trigonometric Functions

    Analyzing the Graphs of y = sec x and y = cscx. The secant was defined by the reciprocal identity sec x = 1 cos x. sec x = 1 cos x. Notice that the function is undefined when the cosine is 0, leading to vertical asymptotes at π 2, π 2, 3 π 2, 3 π 2, etc. Because the cosine is never more than 1 in absolute value, the secant, being the reciprocal, will never be less than 1 in absolute value.

  5. Chapter 7.3: Graphs of the Other Trigonometric Functions

    Using Trigonometric Functions to Solve Real-World Scenarios. Suppose the function marks the distance in the movement of a light beam from the top of a police car across a wall where is the time in seconds and is the distance in feet from a point on the wall directly across from the police car. Find and interpret the stretching factor and period.

  6. Graphs of the Other Trigonometric Functions

    We can graph [latex]\,y=\mathrm {csc}\,x\, [/latex]by observing the graph of the sine function because these two functions are reciprocals of one another. See . The graph of sine is shown as a dashed orange wave so we can see the relationship. Where the graph of the sine function decreases, the graph of the increases.

  7. Trigonometric functions

    This topic covers: - Unit circle definition of trig functions - Trig identities - Graphs of sinusoidal & trigonometric functions - Inverse trig functions & solving trig equations - Modeling with trig functions - Parametric functions Introduction to radians Learn Intro to radians Radians & degrees Degrees to radians Radians to degrees

  8. Graphs of Other Trigonometric Functions

    Learn Graphs of Other Trigonometric Functions with free step-by-step video explanations and practice problems by experienced tutors. Skip to main content. Precalculus Start typing, then use the up and down arrows to select an option from the list. ... Mario's Math Tutoring. 352. views. 05:05. PreCalculus - Trigonometry (26 of 54) Graphing y=tan ...

  9. Graphs of Trigonometric Functions

    To solve the problems on this page, you should be familiar with the following: Sine and Cosine Graphs. Cosec and Sec Graphs. Tangent and Cotangent Graphs. Inverse Trigonometric Graphs. Trigonometric Graphs - Amplitude and Periodicity. Graphical Transformation of Trigonometric Functions.

  10. 2.1: Graphs of the Sine and Cosine Functions

    y = Asin(Bx − C) + D. y = Acos(Bx − C) + D. The graph could represent either a sine or a cosine function that is shifted and/or reflected. When x = 0, the graph has an extreme point, (0, 0). Since the cosine function has an extreme point for x = 0, let us write our equation in terms of a cosine function.

  11. Trigonometric functions

    Learn how to use sine, cosine, and tangent to solve real-world problems involving triangles and circular motion. Unit circle introduction Learn Unit circle The trig functions & right triangle trig ratios Trig unit circle review Trigonometric functions: FAQ Practice Up next for you: Unit circle Get 3 of 4 questions to level up! Start Not started

  12. Solving Real-Life Problems Using Trigonometry

    Introduction to Trigonometry Choosing a Trigonometric Ratio to Use Calculating Angles & Lengths Using Trigonometry Angles of Elevation & Depression 3D Trigonometry Problems Trigonometry & Bearings 2-Minute Feedback Form

  13. Interpreting Trigonometric Graphs in a Real-world Context

    Trigonometry Skills Practice 1. The following graphs shows swing of a ball attached to a crane. The ball moves continuously back and forth. Find the amplitude of the ball using the graph...

  14. Graphs of the Other Trigonometric Functions

    Analyzing the Graph of y = cot x. The last trigonometric function we need to explore is cotangent.The cotangent is defined by the reciprocal identity Notice that the function is undefined when the tangent function is 0, leading to a vertical asymptote in the graph at etc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers.

  15. 5.1: Graphing the Trigonometric Functions

    The graphs of the remaining trigonometric functions can be determined by looking at the graphs of their reciprocal functions. For example, using csc x = 1sin x csc x = 1 sin x we can just look at the graph of y = sin x y = sin x and invert the values. We will get vertical asymptotes when sin x = 0 sin x = 0, namely at multiples of π π: x = 0 ...

  16. Graphs of the Other Trigonometric Functions

    Graphs of the Other Trigonometric Functions Learning Outcomes Analyze the graph of y=tan x and y=cot x. Graph variations of y=tan x and y=cot x. Determine a function formula from a tangent or cotangent graph. Analyze the graphs of y=sec x and y=csc x. Graph variations of y=sec x and y=csc x.

  17. Trigonometric Functions

    The graphs of trigonometric functions have the domain value of θ represented on the horizontal x-axis and the range value represented along the vertical y-axis. The graphs of Sinθ and Tanθ pass through the origin and the graphs of other trigonometric functions do not pass through the origin. The range of Sinθ and Cosθ is limited to [-1, 1].

  18. Exercises: 4.2 Graphs of Trigonometric Functions

    Prepare a graph with the horizontal axis scaled from 0° 0 ° to 360° 360 ° in multiples of 30°. 30 °. θ by plotting points for multiples of 30°. 30 °. Exercise Group. For Problems 27-30, give the coordinates of each point on the graph of f (θ) = sinθ f ( θ) = sin θ or f (θ) = cosθ f ( θ) = cos θ -27. 27.

  19. 2.E: Graphs of the Trigonometric Functions (Exercises)

    2.E: Graphs of the Trigonometric Functions (Exercises) Page ID. Exercise 2.E. 1. In each of the following, the graph on the left shows the terminal point of an arc t (with 0 ≤ t ≤ 2π) on the unit circle. The graphs on the right show the graphs of y = cos(t) and y = sin(t) with some points on the graph labeled.

  20. PDF Supplementary Trigonometry Exercise Problems

    Trig Section 5.1: Graphing the Trigonometric Functions / Unit Circle MULTIPLE CHOICE. Solve the problem. 1) What is the domain of the cosine function? 1) A) all real numbers, except integral multiples of (180 °) B) all real numbers C) all real numbers, except odd multiples of 2 (90 °) D) all real numbers from - 1 to 1, inclusive

  21. Calculus I

    Note that the point of these problems is not really to learn how to find the value of trig functions but instead to get you comfortable with the unit circle since that is a very important skill that will be needed in solving trig equations. cos(5π 6) cos. ⁡. ( 5 π 6) Solution. sin(−4π 3) sin. ⁡.

  22. 2.6: Solving Trigonmetric Equations

    We solved algebraic equations in algebra and now we will solve trigonometric equations. A trigonometric equation is an equation that involves trigonometric functions. We have already used graphical methods to approximate solutions of trigonometric equations. In Example 2.17, we used the function \[V(t) = 35\cos \left(\dfrac{5\pi}{3}t \right ...