- 7.1 Solving Trigonometric Equations with Identities
- 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
- 6.2 Graphs of the Other Trigonometric Functions
- 6.3 Inverse Trigonometric Functions
- Introduction to Trigonometric Identities and Equations
- 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:

- Verify the fundamental trigonometric identities.
- Simplify trigonometric expressions using algebra and the identities.

In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports—there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation.

In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions.

## Verifying the Fundamental Trigonometric Identities

Identities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways.

To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the Pythagorean Identities , the even-odd identities, the reciprocal identities, and the quotient identities.

We will begin with the Pythagorean Identities (see Table 1 ), which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities.

The second and third identities can be obtained by manipulating the first. The identity 1 + cot 2 θ = csc 2 θ 1 + cot 2 θ = csc 2 θ is found by rewriting the left side of the equation in terms of sine and cosine.

Prove: 1 + cot 2 θ = csc 2 θ 1 + cot 2 θ = csc 2 θ

Similarly, 1 + tan 2 θ = sec 2 θ 1 + tan 2 θ = sec 2 θ can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives

The next set of fundamental identities is the set of even-odd identities. The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle and determine whether the identity is odd or even. (See Table 2 ).

Recall that an odd function is one in which f (− x ) = − f ( x ) f (− x ) = − f ( x ) for all x x in the domain of f . f . The sine function is an odd function because sin ( − θ ) = − sin θ . sin ( − θ ) = − sin θ . The graph of an odd function is symmetric about the origin. For example, consider corresponding inputs of π 2 π 2 and − π 2 . − π 2 . The output of sin ( π 2 ) sin ( π 2 ) is opposite the output of sin ( − π 2 ) . sin ( − π 2 ) . Thus,

This is shown in Figure 2 .

Recall that an even function is one in which

The graph of an even function is symmetric about the y- axis. The cosine function is an even function because cos ( − θ ) = cos θ . cos ( − θ ) = cos θ . For example, consider corresponding inputs π 4 π 4 and − π 4 . − π 4 . The output of cos ( π 4 ) cos ( π 4 ) is the same as the output of cos ( − π 4 ) . cos ( − π 4 ) . Thus,

See Figure 3 .

For all θ θ in the domain of the sine and cosine functions, respectively, we can state the following:

- Since sin (− θ ) = − sin θ , sin (− θ ) = − sin θ , sine is an odd function.
- Since, cos (− θ ) = cos θ , cos (− θ ) = cos θ , cosine is an even function.

The other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, consider the tangent identity, tan (− θ ) = −tan θ . tan (− θ ) = −tan θ . We can interpret the tangent of a negative angle as tan (− θ ) = sin ( − θ ) cos (− θ ) = − sin θ cos θ = − tan θ . tan (− θ ) = sin ( − θ ) cos (− θ ) = − sin θ cos θ = − tan θ . Tangent is therefore an odd function, which means that tan ( − θ ) = − tan ( θ ) tan ( − θ ) = − tan ( θ ) for all θ θ in the domain of the tangent function .

The cotangent identity, cot ( − θ ) = − cot θ , cot ( − θ ) = − cot θ , also follows from the sine and cosine identities. We can interpret the cotangent of a negative angle as cot ( − θ ) = cos ( − θ ) sin ( − θ ) = cos θ − sin θ = − cot θ . cot ( − θ ) = cos ( − θ ) sin ( − θ ) = cos θ − sin θ = − cot θ . Cotangent is therefore an odd function, which means that cot ( − θ ) = − cot ( θ ) cot ( − θ ) = − cot ( θ ) for all θ θ in the domain of the cotangent function .

The cosecant function is the reciprocal of the sine function, which means that the cosecant of a negative angle will be interpreted as csc ( − θ ) = 1 sin ( − θ ) = 1 − sin θ = − csc θ . csc ( − θ ) = 1 sin ( − θ ) = 1 − sin θ = − csc θ . The cosecant function is therefore odd.

Finally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted as sec ( − θ ) = 1 cos ( − θ ) = 1 cos θ = sec θ . sec ( − θ ) = 1 cos ( − θ ) = 1 cos θ = sec θ . The secant function is therefore even.

To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities.

The next set of fundamental identities is the set of reciprocal identities , which, as their name implies, relate trigonometric functions that are reciprocals of each other. See Table 3 .

The final set of identities is the set of quotient identities , which define relationships among certain trigonometric functions and can be very helpful in verifying other identities. See Table 4 .

The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.

## Summarizing Trigonometric Identities

The Pythagorean Identities are based on the properties of a right triangle.

The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.

The reciprocal identities define reciprocals of the trigonometric functions.

The quotient identities define the relationship among the trigonometric functions.

## Graphing the Equations of an Identity

Graph both sides of the identity cot θ = 1 tan θ . cot θ = 1 tan θ . In other words, on the graphing calculator, graph y = cot θ y = cot θ and y = 1 tan θ . y = 1 tan θ .

See Figure 4 .

We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to confirm an identity verified with analytical means. If both expressions give the same graph, then they are most likely identities.

Given a trigonometric identity, verify that it is true.

- Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.
- Look for opportunities to factor expressions, square a binomial, or add fractions.
- Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.
- If these steps do not yield the desired result, try converting all terms to sines and cosines.

## Verifying a Trigonometric Identity

Verify tan θ cos θ = sin θ . tan θ cos θ = sin θ .

We will start on the left side, as it is the more complicated side:

This identity was fairly simple to verify, as it only required writing tan θ tan θ in terms of sin θ sin θ and cos θ . cos θ .

Verify the identity csc θ cos θ tan θ = 1. csc θ cos θ tan θ = 1.

## Verifying the Equivalency Using the Even-Odd Identities

Verify the following equivalency using the even-odd identities:

Working on the left side of the equation, we have

## Verifying a Trigonometric Identity Involving sec 2 θ

Verify the identity sec 2 θ − 1 sec 2 θ = sin 2 θ sec 2 θ − 1 sec 2 θ = sin 2 θ

As the left side is more complicated, let’s begin there.

There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.

In the first method, we used the identity sec 2 θ = tan 2 θ + 1 sec 2 θ = tan 2 θ + 1 and continued to simplify. In the second method, we split the fraction, putting both terms in the numerator over the common denominator. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same.

Show that cot θ csc θ = cos θ . cot θ csc θ = cos θ .

## Creating and Verifying an Identity

Create an identity for the expression 2 tan θ sec θ 2 tan θ sec θ by rewriting strictly in terms of sine.

There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:

## Verifying an Identity Using Algebra and Even/Odd Identities

Verify the identity:

Let’s start with the left side and simplify:

Verify the identity sin 2 θ − 1 tan θ sin θ − tan θ = sin θ + 1 tan θ . sin 2 θ − 1 tan θ sin θ − tan θ = sin θ + 1 tan θ .

## Verifying an Identity Involving Cosines and Cotangents

Verify the identity: ( 1 − cos 2 x ) ( 1 + cot 2 x ) = 1. ( 1 − cos 2 x ) ( 1 + cot 2 x ) = 1.

We will work on the left side of the equation.

## Using Algebra to Simplify Trigonometric Expressions

We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations.

For example, the equation ( sin x + 1 ) ( sin x − 1 ) = 0 ( sin x + 1 ) ( sin x − 1 ) = 0 resembles the equation ( x + 1 ) ( x − 1 ) = 0 , ( x + 1 ) ( x − 1 ) = 0 , which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations.

Another example is the difference of squares formula, a 2 − b 2 = ( a − b ) ( a + b ) , a 2 − b 2 = ( a − b ) ( a + b ) , which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve.

## Writing the Trigonometric Expression as an Algebraic Expression

Write the following trigonometric expression as an algebraic expression: 2 cos 2 θ + cos θ − 1. 2 cos 2 θ + cos θ − 1.

Notice that the pattern displayed has the same form as a standard quadratic expression, a x 2 + b x + c . a x 2 + b x + c . Letting cos θ = x , cos θ = x , we can rewrite the expression as follows:

This expression can be factored as ( 2 x − 1 ) ( x + 1 ) . ( 2 x − 1 ) ( x + 1 ) . If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for x . x . At this point, we would replace x x with cos θ cos θ and solve for θ . θ .

## Rewriting a Trigonometric Expression Using the Difference of Squares

Rewrite the trigonometric expression: 4 cos 2 θ − 1. 4 cos 2 θ − 1.

Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares. Thus,

If this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let cos θ = x , cos θ = x , rewrite the expression as 4 x 2 − 1 , 4 x 2 − 1 , and factor ( 2 x − 1 ) ( 2 x + 1 ) . ( 2 x − 1 ) ( 2 x + 1 ) . Then replace x x with cos θ cos θ and solve for the angle.

Rewrite the trigonometric expression: 25 − 9 sin 2 θ . 25 − 9 sin 2 θ .

## Simplify by Rewriting and Using Substitution

Simplify the expression by rewriting and using identities:

We can start with the Pythagorean identity.

Now we can simplify by substituting 1 + cot 2 θ 1 + cot 2 θ for csc 2 θ . csc 2 θ . We have

Use algebraic techniques to verify the identity: cos θ 1 + sin θ = 1 − sin θ cos θ . cos θ 1 + sin θ = 1 − sin θ cos θ .

(Hint: Multiply the numerator and denominator on the left side by 1 − sin θ . ) 1 − sin θ . )

Access these online resources for additional instruction and practice with the fundamental trigonometric identities.

- Fundamental Trigonometric Identities
- Verifying Trigonometric Identities

## 7.1 Section Exercises

We know g ( x ) = cos x g ( x ) = cos x is an even function, and f ( x ) = sin x f ( x ) = sin x and h ( x ) = tan x h ( x ) = tan x are odd functions. What about G ( x ) = cos 2 x , F ( x ) = sin 2 x , G ( x ) = cos 2 x , F ( x ) = sin 2 x , and H ( x ) = tan 2 x ? H ( x ) = tan 2 x ? Are they even, odd, or neither? Why?

Examine the graph of f ( x ) = sec x f ( x ) = sec x on the interval [ − π , π ] . [ − π , π ] . How can we tell whether the function is even or odd by only observing the graph of f ( x ) = sec x ? f ( x ) = sec x ?

After examining the reciprocal identity for sec t , sec t , explain why the function is undefined at certain points.

All of the Pythagorean Identities are related. Describe how to manipulate the equations to get from sin 2 t + cos 2 t = 1 sin 2 t + cos 2 t = 1 to the other forms.

For the following exercises, use the fundamental identities to fully simplify the expression.

sin x cos x sec x sin x cos x sec x

sin ( − x ) cos ( − x ) csc ( − x ) sin ( − x ) cos ( − x ) csc ( − x )

tan x sin x + sec x cos 2 x tan x sin x + sec x cos 2 x

csc x + cos x cot ( − x ) csc x + cos x cot ( − x )

cot t + tan t sec ( − t ) cot t + tan t sec ( − t )

3 sin 3 t csc t + cos 2 t + 2 cos ( − t ) cos t 3 sin 3 t csc t + cos 2 t + 2 cos ( − t ) cos t

− tan ( − x ) cot ( − x ) − tan ( − x ) cot ( − x )

− sin ( − x ) cos x sec x csc x tan x cot x − sin ( − x ) cos x sec x csc x tan x cot x

1 + tan 2 θ csc 2 θ + sin 2 θ + 1 sec 2 θ 1 + tan 2 θ csc 2 θ + sin 2 θ + 1 sec 2 θ

( tan x csc 2 x + tan x sec 2 x ) ( 1 + tan x 1 + cot x ) − 1 cos 2 x ( tan x csc 2 x + tan x sec 2 x ) ( 1 + tan x 1 + cot x ) − 1 cos 2 x

1 − cos 2 x tan 2 x + 2 sin 2 x 1 − cos 2 x tan 2 x + 2 sin 2 x

For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression.

tan x + cot x csc x ; cos x tan x + cot x csc x ; cos x

sec x + csc x 1 + tan x ; sin x sec x + csc x 1 + tan x ; sin x

cos x 1 + sin x + tan x ; cos x cos x 1 + sin x + tan x ; cos x

1 sin x cos x − cot x ; cot x 1 sin x cos x − cot x ; cot x

1 1 − cos x − cos x 1 + cos x ; csc x 1 1 − cos x − cos x 1 + cos x ; csc x

( sec x + csc x ) ( sin x + cos x ) − 2 − cot x ; tan x ( sec x + csc x ) ( sin x + cos x ) − 2 − cot x ; tan x

1 csc x − sin x ; sec x and tan x 1 csc x − sin x ; sec x and tan x

1 − sin x 1 + sin x − 1 + sin x 1 − sin x ; sec x and tan x 1 − sin x 1 + sin x − 1 + sin x 1 − sin x ; sec x and tan x

tan x ; sec x tan x ; sec x

sec x ; cot x sec x ; cot x

sec x ; sin x sec x ; sin x

cot x ; sin x cot x ; sin x

cot x ; csc x cot x ; csc x

For the following exercises, verify the identity.

cos x − cos 3 x = cos x sin 2 x cos x − cos 3 x = cos x sin 2 x

cos x ( tan x − sec ( − x ) ) = sin x − 1 cos x ( tan x − sec ( − x ) ) = sin x − 1

1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = 1 + 2 tan 2 x 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = 1 + 2 tan 2 x

( sin x + cos x ) 2 = 1 + 2 sin x cos x ( sin x + cos x ) 2 = 1 + 2 sin x cos x

cos 2 x − tan 2 x = 2 − sin 2 x − sec 2 x cos 2 x − tan 2 x = 2 − sin 2 x − sec 2 x

For the following exercises, prove or disprove the identity.

1 1 + cos x − 1 1 − cos ( − x ) = − 2 cot x csc x 1 1 + cos x − 1 1 − cos ( − x ) = − 2 cot x csc x

csc 2 x ( 1 + sin 2 x ) = cot 2 x csc 2 x ( 1 + sin 2 x ) = cot 2 x

( sec 2 ( − x ) − tan 2 x tan x ) ( 2 + 2 tan x 2 + 2 cot x ) − 2 sin 2 x = cos 2 x ( sec 2 ( − x ) − tan 2 x tan x ) ( 2 + 2 tan x 2 + 2 cot x ) − 2 sin 2 x = cos 2 x

tan x sec x sin ( − x ) = cos 2 x tan x sec x sin ( − x ) = cos 2 x

sec ( − x ) tan x + cot x = − sin ( − x ) sec ( − x ) tan x + cot x = − sin ( − x )

1 + sin x cos x = cos x 1 + sin ( − x ) 1 + sin x cos x = cos x 1 + sin ( − x )

For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression.

cos 2 θ − sin 2 θ 1 − tan 2 θ = sin 2 θ cos 2 θ − sin 2 θ 1 − tan 2 θ = sin 2 θ

3 sin 2 θ + 4 cos 2 θ = 3 + cos 2 θ 3 sin 2 θ + 4 cos 2 θ = 3 + cos 2 θ

sec θ + tan θ cot θ + cos θ = sec 2 θ sec θ + tan θ cot θ + cos θ = sec 2 θ

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## Trigonometric Identities Problems

Exercise 10, exercise 11, exercise 12, solution of exercise 1, solution of exercise 2, solution of exercise 3, solution of exercise 4, solution of exercise 5, solution of exercise 6, solution of exercise, solution of exercise 7, solution of exercise 8, solution of exercise 9, solution of exercise 10, solution of exercise 11, solution of exercise 12.

PS: Note that all the angles are in degree .

Prove the identities:

Simplify the fractions:

First quadrant:

Second quadrant:

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## Trigonometric Values

Trigonometric identities, trigonometric ratios, trigonometric functions, trigonometric formulas, systems of simultaneous trigonometric equations, unit circle, sine and cosine law, solving right triangles, even-odd identities, oblique triangles, pythagorean identities, measurement of angles, other trigonometric identities, trigonometric equations, cofunction identities, oblique triangle word problems, trigonometric angles worksheet, right triangle word problems, trigonometric equations worksheet, cancel reply.

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## Trigonometry : Solving Word Problems with Trigonometry

Study concepts, example questions & explanations for trigonometry, all trigonometry resources, example questions, example question #1 : solving word problems with trigonometry.

You can draw the following right triangle using the information given by the question:

Since you want to find the height of the platform, you will need to use tangent.

You can draw the following right triangle from the information given by the question.

In order to find the height of the flagpole, you will need to use tangent.

You can draw the following right triangle from the information given in the question:

In order to find out how far up the ladder goes, you will need to use sine.

## Example Question #4 : Solving Word Problems With Trigonometry

In right triangle ABC, where angle A measures 90 degrees, side AB measures 15 and side AC measures 36, what is the length of side BC?

This triangle cannot exist.

## Example Question #5 : Solving Word Problems With Trigonometry

A support wire is anchored 10 meters up from the base of a flagpole, and the wire makes a 25 o angle with the ground. How long is the wire, w? Round your answer to two decimal places.

23.81 meters

28.31 meters

21.83 meters

To make sense of the problem, start by drawing a diagram. Label the angle of elevation as 25 o , the height between the ground and where the wire hits the flagpole as 10 meters, and our unknown, the length of the wire, as w.

Now, we just need to solve for w using the information given in the diagram. We need to ask ourselves which parts of a triangle 10 and w are relative to our known angle of 25 o . 10 is opposite this angle, and w is the hypotenuse. Now, ask yourself which trig function(s) relate opposite and hypotenuse. There are two correct options: sine and cosecant. Using sine is probably the most common, but both options are detailed below.

We know that sine of a given angle is equal to the opposite divided by the hypotenuse, and cosecant of an angle is equal to the hypotenuse divided by the opposite (just the reciprocal of the sine function). Therefore:

To solve this problem instead using the cosecant function, we would get:

The reason that we got 23.7 here and 23.81 above is due to differences in rounding in the middle of the problem.

## Example Question #6 : Solving Word Problems With Trigonometry

When the sun is 22 o above the horizon, how long is the shadow cast by a building that is 60 meters high?

To solve this problem, first set up a diagram that shows all of the info given in the problem.

Next, we need to interpret which side length corresponds to the shadow of the building, which is what the problem is asking us to find. Is it the hypotenuse, or the base of the triangle? Think about when you look at a shadow. When you see a shadow, you are seeing it on something else, like the ground, the sidewalk, or another object. We see the shadow on the ground, which corresponds to the base of our triangle, so that is what we'll be solving for. We'll call this base b.

Therefore the shadow cast by the building is 150 meters long.

If you got one of the incorrect answers, you may have used sine or cosine instead of tangent, or you may have used the tangent function but inverted the fraction (adjacent over opposite instead of opposite over adjacent.)

## Example Question #7 : Solving Word Problems With Trigonometry

From the top of a lighthouse that sits 105 meters above the sea, the angle of depression of a boat is 19 o . How far from the boat is the top of the lighthouse?

423.18 meters

318.18 meters

36.15 meters

110.53 meters

To solve this problem, we need to create a diagram, but in order to create that diagram, we need to understand the vocabulary that is being used in this question. The following diagram clarifies the difference between an angle of depression (an angle that looks downward; relevant to our problem) and the angle of elevation (an angle that looks upward; relevant to other problems, but not this specific one.) Imagine that the top of the blue altitude line is the top of the lighthouse, the green line labelled GroundHorizon is sea level, and point B is where the boat is.

Merging together the given info and this diagram, we know that the angle of depression is 19 o and and the altitude (blue line) is 105 meters. While the blue line is drawn on the left hand side in the diagram, we can assume is it is the same as the right hand side. Next, we need to think of the trig function that relates the given angle, the given side, and the side we want to solve for. The altitude or blue line is opposite the known angle, and we want to find the distance between the boat (point B) and the top of the lighthouse. That means that we want to determine the length of the hypotenuse, or red line labelled SlantRange. The sine function relates opposite and hypotenuse, so we'll use that here. We get:

## Example Question #8 : Solving Word Problems With Trigonometry

Angelina just got a new car, and she wants to ride it to the top of a mountain and visit a lookout point. If she drives 4000 meters along a road that is inclined 22 o to the horizontal, how high above her starting point is she when she arrives at the lookout?

9.37 meters

1480 meters

3708.74 meters

10677.87 meters

1616.1 meters

As with other trig problems, begin with a sketch of a diagram of the given and sought after information.

Angelina and her car start at the bottom left of the diagram. The road she is driving on is the hypotenuse of our triangle, and the angle of the road relative to flat ground is 22 o . Because we want to find the change in height (also called elevation), we want to determine the difference between her ending and starting heights, which is labelled x in the diagram. Next, consider which trig function relates together an angle and the sides opposite and hypotenuse relative to it; the correct one is sine. Then, set up:

Therefore the change in height between Angelina's starting and ending points is 1480 meters.

## Example Question #9 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 50 feet apart. The shorter building is 40 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 48 o . How high is the taller building?

To solve this problem, let's start by drawing a diagram of the two buildings, the distance in between them, and the angle between the tops of the two buildings. Then, label in the given lengths and angle.

## Example Question #10 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 80 feet apart. The shorter building is 55 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 32 o . How high is the taller building?

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## PROBLEMS ON TRIGONOMETRIC IDENTITIES WITH SOLUTIONS

Problem 1 :

Prove :

(1 - cos 2 θ) csc 2 θ = 1

Let A = (1 - cos 2 θ) csc 2 θ and B = 1.

A = (1 - cos 2 θ)csc 2 θ

Because sin 2 θ + cos 2 θ = 1, we have

sin 2 θ = 1 - cos 2 θ

Then,

A = sin 2 θ ⋅ csc 2 θ

A = sin 2 θ ⋅ (1/sin 2 θ)

A = sin 2 θ /sin 2 θ

A = 1

A = B (Proved)

Problem 2 :

sec θ √(1 - sin 2 θ) = 1

Let A = sec θ √(1 - sin 2 θ) and B = 1.

A = sec θ √(1 - sin 2 θ)

cos 2 θ = 1 - sin 2 θ

A = sec θ √cos 2 θ

A = sec θ ⋅ cos θ

A = sec θ ⋅ (1/sec θ)

A = sec θ / sec θ

Problem 3 :

tan θ sin θ + cos θ = sec θ

Let A = tan θ sin θ + cos θ and B = sec θ.

A = tan θ sin θ + cos θ

A = (sin θ/cos θ) ⋅ sin θ + cos θ

A = (sin 2 θ /cos θ) + cos θ

A = (sin 2 θ /cos θ) + (cos 2 θ/cosθ)

A = (sin 2 θ + cos 2 θ ) / cos θ

A = 1 / cos θ

A = sec θ

Problem 4 :

(1 - cos θ)(1 + cos θ)(1 + cot 2 θ) = 1

Let A = (1 - cos θ)(1 + cos θ)(1 + cot 2 θ) = 1 and B = 1.

A = (1 - cos θ)(1 + cos θ)(1 + cot 2 θ)

A = (1 - cos 2 θ)(1 + cot 2 θ)

A = sin 2 θ ⋅ (1 + cot 2 θ)

A = sin 2 θ + sin 2 θ ⋅ cot 2 θ

A = sin 2 θ + sin 2 θ ⋅ (cos 2 θ/sin 2 θ)

A = sin 2 θ + cos 2 θ

Problem 5 :

cot θ + tan θ = sec θ csc θ

Let A = cot θ + tan θ and B = sec θ csc θ.

A = cot θ + tan θ

A = (cos θ/sin θ) + (sin θ/cos θ)

A = (cos 2 θ/sin θ cos θ) + (sin 2 θ/sin θ cos θ)

A = (cos 2 θ + sin 2 θ) / sin θ cos θ

A = 1 / sin θ cos θ

A = (1/cos θ) ⋅ (1/sin θ)

A = sec θ csc θ

Problem 6 :

cos θ/(1 - tan θ) + sin θ/(1 - cot θ) = sin θ + cos θ

Let A = cos θ/(1 - tan θ) + sin θ/(1 - cot θ) and

B = sin θ + cos θ

A = cos θ/{1 - (sin θ/cos θ)} + sin θ/{1 - (cos θ/sin θ)}

A = cos 2 θ /(cos θ - sin θ) + sin 2 θ /(sin θ - cos θ)

A = cos 2 θ /(cos θ - sin θ) - sin 2 θ /(cos θ - sin θ)

A = (cos 2 θ - sin 2 θ ) / (cos θ - sin θ)

A = [(cos θ + sin θ)(cos θ - sin θ)] / (cos θ - sin θ)

A = (cos θ + sin θ)

Problem 7 :

tan 4 θ + tan 2 θ = sec 4 θ - sec 2 θ

Let A = tan 4 θ + tan 2 θ and B = sec 4 θ + sec 2 θ.

A = tan 4 θ + tan 2 θ

A = tan 2 θ (tan 2 θ + 1)

We know that,

tan 2 θ = sec 2 θ - 1

tan 2 θ + 1 = sec 2 θ

A = (sec 2 θ - 1)(sec 2 θ )

A = sec 4 θ - sec 2 θ

Problem 8 :

√{(sec θ – 1)/(sec θ + 1)} = cosec θ - cot θ

Let A = √{(sec θ – 1)/(sec θ + 1)} and B = cosec θ - cot θ.

A = √{(sec θ – 1)/(sec θ + 1)}

A = √[{(sec θ - 1) (sec θ - 1)}/{(sec θ + 1) (sec θ - 1)}]

A = √{(sec θ - 1) 2 / (sec 2 θ - 1)}

A = √{(sec θ - 1) 2 / tan 2 θ }

A = (sec θ – 1)/tan θ

A = (sec θ/tan θ) – (1/tan θ)

A = {(1/cos θ)/(sin θ/cos θ)} - cot θ

A = {(1/cos θ) ⋅ (cos θ/sin θ)} - cot θ

A = (1/sin θ) - cot θ

A = cosec θ - cot θ

Problem 9 :

(1 - sin A)/(1 + sin A) = (sec A - tan A) 2

Let A = (1 - sin A)/(1 + sin A) and B = (sec A - tan A) 2 .

A = (1 - sin A) / (1 + sin A)

A = (1 - sin A) 2 / (1 - sin A) (1 + sin A)

A = (1 - sin A) 2 / (1 - sin 2 A)

A = (1 - sin A) 2 / (cos 2 A)

A = (1 - sin A) 2 / (cos A) 2

A = {(1 - sin A) / cos A} 2

A = {(1/cos A) - (sin A/cos A)} 2

A = (sec A – tan A) 2

Problem 10 :

(tan θ + sec θ - 1)/(tan θ - sec θ + 1) = (1 + sin θ)/cos θ

Let A = (tan θ + sec θ - 1)/(tan θ - sec θ + 1) and

B = (1 + sin θ)/cos θ.

A = (tan θ + sec θ - 1)/(tan θ - sec θ + 1)

A = [(tan θ + sec θ) - (sec 2 θ - tan 2 θ )]/(tan θ - sec θ + 1)

A = {(tan θ + sec θ) (1 - sec θ + tan θ)}/(tan θ - sec θ + 1)

A = {(tan θ + sec θ) (tan θ - sec θ + 1)}/(tan θ - sec θ + 1)

A = tan θ + sec θ

A = (sin θ/cos θ) + (1/cos θ)

A = (sin θ + 1)/cos θ

A = (1 + sin θ)/cos θ

A = B (Proved)

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- Problems on Trigonometric Identities

Here we will prove the problems on trigonometric identities. In an identity there are two sides of the equation, one side is known as ‘left hand side’ and the other side is known as ‘right hand side’ and to prove the identity we need to use logical steps showing that one side of the equation ends up with the other side of the equation.

Proving the problems on trigonometric identities:

● Trigonometric Functions

- Basic Trigonometric Ratios and Their Names
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