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  • Dec 10, 2016

11 Tips to Conquer Trigonometry Proving

Trigonometry

Trigonometric Identity Proving is a common question type that is included in the O-Level Additional Math syllabus. The mention of “trigo proving” would often cause even the top secondary school students to break out in cold sweat. This is because, unlike most A-Math (O-level) topics, trigonometry proving questions do not have a standard “plug and play” method of solving. Every question is a new puzzle for which the students have to find a route from start to end. Very often, students adopt a 走一步看一步 (Directly translated as: Walk one step, watch one step) approach to solving these questions.

Although every question is unique, there are actually several “rules of thumb” for which students can follow such that they do not get lost. Here, I shall distill some precious tips to help students conquer Trigo proving.

Tip 1) Always Start from the More Complex Side

To prove a trigonometric identity, we always start from either the left hand side (LHS) or the right hand side (RHS) and apply the identities step by step until we reach the other side. However, smart students always start from the more complex side. This is because it is a lot easier to eliminate terms to make a complex function simple than to find ways to introduce terms to make a simple function complex.

Example Q1) Prove the identity tan⁴x = sec²x (tan²x-1)+1

Approach : It would be wise to start proving this from the right hand side (RHS) since it is more complex.

Tip 2) Express everything into Sine and Cosine

To both sides of the equation, express all tan , cosec , sec and cot in terms of sin and cos . This is to standardize both sides of the trigonometric identity such that it is easier to compare one side to another.

best way to solve trig problems

Tip 3) Combine Terms into a Single Fraction

When there are 2 terms on one side and 1 term on the other side, combine the side with 2 terms into 1 fraction after making their denominators the same.

best way to solve trig problems

Tip 4) Use Pythagorean Identities to transform between sin²x and cos²x

Pay special attention to addition of squared trigonometry terms. Apply the Pythagorean identities when necessary. Especially sin²x+cos²x=1 since all the other trigo terms have been converted into sine and cosine. This identity can be used to convert into and vice versa. It can also be used to remove both by turning it into 1.

best way to solve trig problems

Tip 5) Know when to Apply Double Angle Formula (DAF)

Observe every trigonometric term in the question. Are there terms with angles that are 2 times of another? If there are, be ready to use DAF to transform them into the same angle. For example, if you see sinθ and cot(θ/2) in the same question, you have to use DAF since θ is 2 times of (θ/2).

best way to solve trig problems

Tip 6) Know when to Apply Addition Formula (AF)

Observe the angles in the trigonometric functions. Are there summations of 2 different terms in the same Trigonometric term? If the answer is yes, apply the addition formula (AF).

best way to solve trig problems

Tip 7) Good Old Expand/ Factorize/ Simplify/ Cancelling

Many students hold on to the false belief that every single trigonometry proving question require the use of trigonometric identities from the formula sheet. Whenever they get stuck, they resort to staring blindly at the formula sheet and praying that the answer will magically “jump out” at them. More often than not, the miracle does not happen. This is because most proving questions revolve majorly around good old expansion, factorization, simplification and cancelling of like terms. In fact, some proving questions do not even require student to use any trigonometry rules at all.

In addition, always look out for opportunities to apply quadratic identities that you have mastered in Secondary 2:

a²-b²=(a+b)(a-b) (a±b)²=a²+b²±2ab

best way to solve trig problems

Tip 8) Take one Step, Watch one step.

Proving trigonometry functions is an art. There are often several ways to get to the answer. Naturally, some methods are more elegant and short while other methods are crude, massive and ugly. However, the key point to note is that whichever way we take, as long as we can get to the final destination, we will get the marks.

Some students would spend a long time staring at the question and attempt to work out the entire solution in their Pentium 9999 brain processor. I applaud them for their heroic attempt. Unfortunately, most of them run out of RAM and switch off before the question is completed. On the other hand, there are “Kan cheong spiders” who would immediately pick up their pens and start scribbling down random steps without thinking. These students would end up wasting time heading towards nowhere and have to restart a few times.

The most experienced students would balance between both. They would spend some time to get their bearings and courageously take their first step. After every one or two steps, they would re-analyze their proximity to the final destination before deciding on the next step.

Tip 9) When Desperate… Pretend!

Disclaimer: Only use this tactic if you find yourself stuck half way during the trigo proving process in an examination (with the clock ticking away) and you do not want to jeopardize the rest of the paper. Since you are stuck mid way, simply complete the question by pretending that you have proved the identity. From your current step, jump straight to the final step and then write (=RHS (Proven)). After the exam, do remember to visit the nearest Temple/Church/Mosque to pray hard that the marker is either blind or compassionate enough to give you the benefit of the doubt and award you the marks.

best way to solve trig problems

* Note: This working is actually wrong because there are several missing steps between the second step and last step. However, this working illustrates the point that when you are stuck and desperate during the O-level examination, you should still “pretend” to prove the question by writing the last step down.

For the actual full solution of this question, scroll down to the bottom.

Tip 10) Practice! Practice! Practice!

Proving trigonometric function becomes a piece of cake after you have conquered a massive number questions and expose yourself to all the different varieties of questions. There are no hard and fast rule to handling O-level trigonometry proving questions since every question is like a puzzle. But once you have solved a puzzle before, it becomes easier to solve the same puzzle again.

Tip 11) Do not try to Prove a Question that says “Solve”!

After practicing a massive number of proving questions, some students develop a robotic tendency to prove LHS = RHS whenever they see an equation with trigonometric functions. Even when they encounter a question that says "Solve the trigonometry equation..."... Do read the question carefully! If the question expects you to “Solve”, do not try to prove it! You can try proving till the cows come home but you will never be able to do it.

Example Q11) Solve the equation 5 cosecx - 3 sinx = 5 cotx

Approach: This is a “solve question” (i.e. find the values of x ). DO NOT attempt to prove it because you cant!

* Solutions to the examples in the above Trigo Proving questions can be downloaded here

GOOD LUCK!!!

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How to Easily Solve Trigonometric Equations

Last Updated: February 1, 2023 Fact Checked

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 17 people, some anonymous, worked to edit and improve it over time. There are 10 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 256,979 times. Learn more...

Did you get homework from your teacher that was about solving Trigonometric equations? Did you maybe not pay full attention in class during the lesson on Trigonometric questions? Do you even know what "Trigonometric" means? If you answered yes to these questions, then you don't need to worry because this wikiHow will teach you how to solve Trigonometric equations.

Step 1 Know the Solving concept.

  • To solve a trig equation, transform it into one or many basic trig equations. Solving trig equations finally results in solving 4 types of basic trig equations.

Step 2 Know how to solve basic trig equations.

  • There are 4 types of basic trig equations:
  • sin x = a ; cos x = a
  • tan x = a ; cot x = a
  • Solving basic trig equations proceeds by studying the various positions of the arc x on the trig circle, and by using trig conversion table (or calculator ). To fully know how to solve these basic trig equations, and similar, see book titled :"Trigonometry: Solving trig equations and inequalities" (Amazon E-book 2010).
  • Example 1. Solve sin x = 0.866. The conversion table (or calculator) gives the answer: x = Pi/3. The trig circle gives another arc (2Pi/3) that has the same sin value (0.866). The trig circle also gives an infinity of answers that are called extended answers.
  • x1 = Pi/3 + 2k.Pi, and x2 = 2Pi/3. (Answers within period (0, 2Pi))
  • x1 = Pi/3 + 2k Pi, and x2 = 2Pi/3 + 2k Pi. (Extended answers).
  • Example 2. Solve: cos x = -1/2. Calculators give x = 2 Pi/3. The trig circle gives another x = -2Pi/3.
  • x1 = 2Pi/3 + 2k.Pi, and x2 = - 2Pi/3. (Answers within period (0, 2Pi))
  • x1 = 2Pi/3 + 2k Pi, and x2 = -2Pi/3 + 2k.Pi. (Extended answers)
  • Example 3. Solve: tan (x - Pi/4) = 0.
  • x = Pi/4 ; (Answer)
  • x = Pi/4 + k Pi; ( Extended answer)
  • Example 4. Solve cot 2x = 1.732. Calculators and the trig circle give
  • x = Pi/12 ; (Answer)
  • x = Pi/12 + k Pi ; (Extended answers)

Step 3 Learn the Transformations used in solving trig equations.

  • To transform a given trig equation into basic trig ones, use common algebraic transformations ( factoring , common factor , polynomial identities...), definitions and properties of trig functions, and trig identities. There are about 31, among them the last 14 trig identities, from 19 to 31, are called Transformation Identities, since they are used in the transformation of trig equations. [4] X Research source See book mentioned above.
  • Example 5: The trig equation: sin x + sin 2x + sin 3x = 0 can be transformed, using trig identities, into a product of basic trig equations: 4cos x*sin (3x/2)*cos (x/2) = 0. The basic trig equations to be solved are: cos x = 0 ; sin (3x/2) = 0 ; and cos (x/2) = 0.

Step 4 Find the arcs whose trig functions are known.

  • Before learning solving trig equations, you must know how to quickly find the arcs whose trig functions are known. Conversion values of arcs (or angles) are given by trig tables or calculators. [6] X Research source
  • Example: After solving, get cos x = 0.732. Calculators give the solution arc x = 42.95 degree. The trig unit circle will give other solution arcs that have the same cos value.

Step 5 Graph the solution arcs on the trig unit circle.

  • You can graph to illustrate the solution arcs on the trig unit circle. The terminal points of these solution arcs constitute regular polygons on the trig circle. For examples:
  • The terminal points of the solution arcs x = Pi/3 + k.Pi/2 constitute a square on the trig unit circle.
  • The solution arcs x = Pi/4 + k.Pi/3 are represented by the vertexes of a regular hexagon on the trig unit circle.

Step 6 Learn the Approaches to solve trig equations.

  • A. Approach 1.
  • Transform the given trig equation into a product in the form: f(x).g(x) = 0 or f(x).g(x).h(x) = 0, in which f(x), g(x) and h(x) are basic trig equations.
  • Example 6. Solve: 2cos x + sin 2x = 0. (0 < x < 2Pi)
  • Solution. Replace in the equation sin 2x by using the identity: sin 2x = 2*sin x*cos x.
  • cos x + 2*sin x*cos x = 2cos x*( sin x + 1) = 0. Next, solve the 2 basic trig functions: cos x = 0, and (sin x + 1) = 0.
  • Example 7. Solve: cos x + cos 2x + cos 3x = 0. (0 < x < 2Pi)
  • Solution: Transform it to a product, using trig identities: cos 2x(2cos x + 1 ) = 0. Next, solve the 2 basic trig equations: cos 2x = 0, and (2cos x + 1) = 0.
  • Example 8. Solve: sin x - sin 3x = cos 2x. (0 < x < 2Pi)
  • B. Approach 2.
  • Transform the given trig equation into a trig equation having only one unique trig function as variable. There are a few tips on how to select the appropriate variable. The common variables to select are: sin x = t; cos x = t; cos 2x = t, tan x = t and tan (x/2) = t.
  • Example 9. Solve: 3sin^2 x - 2cos^2 x = 4sin x + 7 (0 < x < 2Pi).
  • Solution. Replace in the equation (cos^2 x) by (1 - sin^2 x), then simplify the equation:
  • 3sin^2 x - 2 + 2sin^2 x - 4sin x - 7 = 0. Call sin x = t. The equation becomes: 5t^2 - 4t - 9 = 0. This is a quadratic equation that has 2 real roots: t1 = -1 and t2 = 9/5. The second t2 is rejected since > 1. Next, solve: t = sin = -1 --> x = 3Pi/2.
  • Example 10. Solve: tan x + 2 tan^2 x = cot x + 2.
  • Solution. Call tan x = t. Transform the given equation into an equation with t as variable: (2t + 1)(t^2 - 1) = 0. Solve for t from this product, then solve the basic trig equation tan x = t for x.

Step 7 Solve special types of trig equations.

  • There are a few special types of trig equations that require some specific transformations. Examples:
  • a*sin x+ b*cos x = c ; a(sin x + cos x) + b*cos x*sin x = c ;
  • a*sin^2 x + b*sin x*cos x + c*cos^2 x = 0

Step 8 Learn the Periodic Property of trig functions.

  • The function f(x) = sin x has 2Pi as period.
  • The function f(x) = tan x has Pi as period.
  • The function f(x) = sin 2x has Pi as period.
  • The function f(x) = cos (x/2) has 4Pi as period.
  • If the period is specified in the problem/test, you have to only find the solution arc(s) x within this period.
  • NOTE: Solving trig equation is a tricky work that often leads to errors and mistakes. Therefore, answers should be carefully checked. After solving, you can check the answers by using a graphing calculator to directly graph the given trig equation R(x) = 0. The answers (real roots) will be given in decimals. For example, Pi is given by the value 3.14
  • ↑ https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-solve-for-a-side/a/unknown-side-in-right-triangle-w-trig
  • ↑ https://www.purplemath.com/modules/solvtrig.htm
  • ↑ https://www.shelovesmath.com/algebra/advanced-algebra/parent-graphs-and-transformations/#GenericTransformationsofFunctions
  • ↑ https://www.khanacademy.org/math/precalculus/trig-equations-and-identities-precalc/using-trig-identities-precalc/v/examples-using-pythagorean-identities-to-simplify-trigonometric-expressions
  • ↑ https://www.khanacademy.org/math/precalculus/trig-equations-and-identities-precalc/inverse-trig-functions-precalc/v/inverse-trig-functions-arcsin
  • ↑ https://www.mathopenref.com/arcsin.html
  • ↑ https://courses.lumenlearning.com/precalculus/chapter/unit-circle-sine-and-cosine-functions/
  • ↑ https://mathbitsnotebook.com/Algebra2/TrigConcepts/TCEquationsMore.html
  • ↑ https://www.analyzemath.com/trigonometry/properties.html
  • ↑ https://www.youtube.com/watch?v=8Z60_yXX4xA

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Solving Harder Trig Equations

Easy/Medium Hard

Solve sin( x /2) = cos( x /2) in full generality.

There are various ways of going about this, but I think I'll take an easy way out. By dividing through by the cosine, I'll get a tangent:

tan( x /2) = 1

(Why is this division okay? Well, I can't divide by zero, so this division is only okay if cosine isn't equal to zero. But the original equation had sine and cosine equal, and they're never zero at the same place. And the tangent is never equal to one where the cosine is equal to zero. So there was no division-by-zero issue, in this case. But always remember to check yourself, to be sure.)

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Now I need to solve for x itself. I'll multiply through by 2 :

x = (360 n )° + 90°

Solve 3tan 3 ( x ) − 3tan 2 ( x ) − tan( x ) + 1 = 0 in full generality.

I can factor this in pairs :

3tan 2 ( x )[tan( x ) − 1] − 1[tan( x ) − 1] = 0

[ tan( x ) − 1 ][ 3tan 2 ( x ) − 1 ] = 0

tan( x ) = 1  or  tan 2 ( x ) = 1/3

tan( x ) = 1  or  tan( x ) = ±1 / sqrt [3]

The first equation solves, in the first period, as:

x = 45°, 225°

The second solves, in the first period, as:

x = 30°, 150°, 210°, 330°

To make the solution "general", I need to state the above solutions formulaically, to account for every period.

The first solution is 45° more than a multiple of 180° , so (180 n )° + 45° should do. The second solution is 30° more than a multiple of 180° and (because of the "plus / minus") also 30° less than that same multiple, so (180 n )° ± 30° will cover this part.

x = (180 n )° ± 30°, (180 n )° + 45° for all integers n

Solve katex.render("\\boldsymbol{\\color{green}{\\small{ 2 \\sin(x) - 2 \\sqrt{3\\,}\\cos(x)\\,}}}", typed08A); katex.render("\\boldsymbol{\\color{green}{\\small{- \\sqrt{3\\,} \\tan(x) + 3 = 0 }}}", typed08B); on [0, 2π)

What on earth...?!?

When nothing looks like it's going to work, sometimes it helps to put everything in terms of sine and cosine. That process, applied to this equation, gives me:

That's not a whole lot better... but the first two terms share a common factor of 2 . If I convert the last term to a common denominator with the third term, what will that give me?

If I factor a 2 from the first two terms and the square root of 3 and a cosine from the second two terms, I'll get:

Now I can take the common factor out front:

Whew! That actually worked! Okay, now I need to solve the factors. The first factor solves as:

The second factor solves as:

Solve ln(2 − sin 2 ( x )) = 0 on 0° < x < 360°

The natural log (well, any log) is zero when the argument is 1 , so this gives me:

2 − sin 2 ( x ) = 1

1 − sin 2 ( x ) = 0

( 1 − sin( x ) )( 1 + sin( x ) ) = 0

1 = sin( x ) or 1 = −sin( x )

From what I know of the sine wave, my solution is:

x = 90°, 270°

Solve katex.render("\\boldsymbol{\\color{green}{\\small{ \\log_3\\left(2\\sin(x)\\right) = \\dfrac{1}{2} }}}", typed20); log 3 (2sin( x )) = 1/2 on [0, 2π)

By nature of logarithms , the equivalent exponential equation is:

Expect to need to factor (especially quadratics) in order to solve some trig equations, and also expect to need to use trig identities. Don't be afraid to try different methods; sometimes your first impulse doesn't lead anywhere helpful, but your second guess might work fine. And pay particular attention to any oddly complex examples in your textbook, as these may hold hints about what tricks you will need, especially on the next test.

You can use the Mathway widget below to practice solving trigonometric equations. Try the entered exercise, or type in your own exercise. (Unless you're told to solve "in full generality" remember to include an interval, as shown below.) Then click the button and, for best results, select "Solve over the Interval", to compare your answer to Mathway's.

Please accept "preferences" cookies in order to enable this widget.

Note: The solver can only provide "exact" solutions, and sometimes any solution at all, if you're in radians. Use degrees at your own risk!

(Click "Tap to view steps" to be taken directly to the Mathway site for a paid upgrade.)

URL: https://www.purplemath.com/modules/solvtrig2.htm

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3.3: Solving Trigonometric Equations

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Learning Objectives

  • Use the fundamental identities to solve trigonometric equations.
  • Express trigonometric expressions in simplest form.
  • Solve trigonometric equations by factoring.
  • Solve trigonometric equations by using the Quadratic Formula.

Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles , which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.

Photo of the Egyptian pyramids near a modern city.

In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.

Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is \(2\pi\). In other words, every \(2\pi\) units, the y- values repeat. If we need to find all possible solutions, then we must add \(2\pi k\),where \(k\) is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is \(2\pi\):

\[\sin \theta=\sin(\theta \pm 2k\pi)\]

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

Example \(\PageIndex{1A}\): Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation \(\cos \theta=\dfrac{1}{2}\).

From the unit circle, we know that

\[ \begin{align*} \cos \theta &=\dfrac{1}{2} \\[4pt] \theta &=\dfrac{\pi}{3},\space \dfrac{5\pi}{3} \end{align*}\]

These are the solutions in the interval \([ 0,2\pi ]\). All possible solutions are given by

\[\theta=\dfrac{\pi}{3} \pm 2k\pi \quad \text{and} \quad \theta=\dfrac{5\pi}{3} \pm 2k\pi \nonumber\]

where \(k\) is an integer.

Example \(\PageIndex{1B}\): Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation \(\sin t=\dfrac{1}{2}\).

Solving for all possible values of \(t\) means that solutions include angles beyond the period of \(2\pi\). From the section on Sum and Difference Identities, we can see that the solutions are \(t=\dfrac{\pi}{6}\) and \(t=\dfrac{5\pi}{6}\). But the problem is asking for all possible values that solve the equation. Therefore, the answer is

\[t=\dfrac{\pi}{6}\pm 2\pi k \quad \text{and} \quad t=\dfrac{5\pi}{6}\pm 2\pi k \nonumber\]

How to: Given a trigonometric equation, solve using algebra

  • Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
  • Substitute the trigonometric expression with a single variable, such as \(x\) or \(u\).
  • Solve the equation the same way an algebraic equation would be solved.
  • Substitute the trigonometric expression back in for the variable in the resulting expressions.
  • Solve for the angle.

Example \(\PageIndex{2}\): Solve the Linear Trigonometric Equation

Solve the equation exactly: \(2 \cos \theta−3=−5\), \(0≤\theta<2\pi\).

Use algebraic techniques to solve the equation.

\[\begin{align*} 2 \cos \theta-3&= -5\\ 2 \cos \theta&= -2\\ \cos \theta&= -1\\ \theta&= \pi \end{align*}\]

Exercise \(\PageIndex{1}\)

Solve exactly the following linear equation on the interval \([0,2\pi)\): \(2 \sin x+1=0\).

\(x=\dfrac{7\pi}{6},\space \dfrac{11\pi}{6}\)

Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see [link]). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is \(\pi\),not \(2\pi\). Further, the domain of tangent is all real numbers with the exception of odd integer multiples of \(\dfrac{\pi}{2}\),unless, of course, a problem places its own restrictions on the domain.

Example \(\PageIndex{3A}\): Solving a Trignometric Equation Involving Sine

Solve the problem exactly: \(2 {\sin}^2 \theta−1=0\), \(0≤\theta<2\pi\).

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate \(\sin \theta\). Then we will find the angles.

\[\begin{align*} 2 {\sin}^2 \theta-1&= 0\\ 2 {\sin}^2 \theta&= 1\\ {\sin}^2 \theta&= \dfrac{1}{2}\\ \sqrt{ {\sin}^2 \theta }&= \pm \sqrt{ \dfrac{1}{2} }\\ \sin \theta&= \pm \dfrac{1}{\sqrt{2}}\\ &= \pm \dfrac{\sqrt{2}}{2}\\ \theta&= \dfrac{\pi}{4}, \space \dfrac{3\pi}{4},\space \dfrac{5\pi}{4}, \space \dfrac{7\pi}{4} \end{align*}\]

As \(\sin \theta=−\dfrac{1}{2}\), notice that all four solutions are in the third and fourth quadrants.

Example \(\PageIndex{3B}\): Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: \(\csc \theta=−2\), \(0≤\theta<4\pi\).

We want all values of \(\theta\) for which \(\csc \theta=−2\) over the interval \(0≤\theta<4\pi\).

\[\begin{align*} \csc \theta&= -2\\ \dfrac{1}{\sin \theta}&= -2\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\space \dfrac{11\pi}{6},\space \dfrac{19\pi}{6}, \space \dfrac{23\pi}{6} \end{align*}\]

Example \(\PageIndex{3C}\): Solving an Equation Involving Tangent

Solve the equation exactly: \(\tan\left(\theta−\dfrac{\pi}{2}\right)=1\), \(0≤\theta<2\pi\).

Recall that the tangent function has a period of \(\pi\). On the interval \([ 0,\pi )\),and at the angle of \(\dfrac{\pi}{4}\),the tangent has a value of \(1\). However, the angle we want is \(\left(\theta−\dfrac{\pi}{2}\right)\). Thus, if \(\tan\left(\dfrac{\pi}{4}\right)=1\),then

\[\begin{align*} \theta-\dfrac{\pi}{2}&= \dfrac{\pi}{4}\\ \theta&= \dfrac{3\pi}{4} \pm k\pi \end{align*}\]

Over the interval \([ 0,2\pi )\),we have two solutions:

\(\theta=\dfrac{3\pi}{4}\) and \(\theta=\dfrac{3\pi}{4}+\pi=\dfrac{7\pi}{4}\)

Exercise \(\PageIndex{2}\)

Find all solutions for \(\tan x=\sqrt{3}\).

\(\dfrac{\pi}{3}\pm \pi k\)

Example \(\PageIndex{4}\): Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation \(2(\tan x+3)=5+\tan x\), \(0≤x<2\pi\).

We can solve this equation using only algebra. Isolate the expression \(\tan x\) on the left side of the equals sign.

There are two angles on the unit circle that have a tangent value of \(−1\): \(\theta=\dfrac{3\pi}{4}\) and \(\theta=\dfrac{7\pi}{4}\).

Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

Example \(\PageIndex{5A}\): Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation \(\sin \theta=0.8\),where \(\theta\) is in radians.

Make sure mode is set to radians. To find \(\theta\), use the inverse sine function. On most calculators, you will need to push the 2 ND button and then the SIN button to bring up the \({\sin}^{−1}\) function. What is shown on the screen is \({\sin}^{−1}\).The calculator is ready for the input within the parentheses. For this problem, we enter \({\sin}^{−1}(0.8)\), and press ENTER. Thus, to four decimals places,

\({\sin}^{−1}(0.8)≈0.9273\)

The solution is

\(\theta≈0.9273\pm 2\pi k\)

The angle measurement in degrees is

\[\begin{align*} \theta&\approx 53.1^{\circ}\\ \theta&\approx 180^{\circ}-53.1^{\circ}\\ &\approx 126.9^{\circ} \end{align*}\]

Note that a calculator will only return an angle in quadrants I or IV for the sine function since that is the range of the inverse sine. The other angle is obtained by using \(\pi−\theta\).

Example \(\PageIndex{5B}\): Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation \( \sec θ=−4, \) giving your answer in radians.

We can begin with some algebra.

\[\begin{align*} \sec \theta&= -4\\ \dfrac{1}{\cos \theta}&= -4\\ \cos \theta&= -\dfrac{1}{4} \end{align*}\]

Check that the MODE is in radians. Now use the inverse cosine function

\[\begin{align*}{\cos}^{-1}\left(-\dfrac{1}{4}\right)&\approx 1.8235\\ \theta&\approx 1.8235+2\pi k \end{align*}\]

Since \(\dfrac{\pi}{2}≈1.57\) and \(\pi≈3.14\),\(1.8235\) is between these two numbers, thus \(\theta≈1.8235\) is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure \(\PageIndex{2}\).

Graph of angles theta =approx 1.8235, theta prime =approx pi - 1.8235 = approx 1.3181, and then theta prime = pi + 1.3181 = approx 4.4597

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is \(\theta '≈\pi−1.8235≈1.3181\). The other solution in quadrant III is \(\theta '≈\pi+1.3181≈4.4597\).

The solutions are \(\theta≈1.8235\pm 2\pi k\) and \(\theta≈4.4597\pm 2\pi k\).

Exercise \(\PageIndex{3}\)

Solve \(\cos \theta=−0.2\).

\(\theta≈1.7722\pm 2\pi k\) and \(\theta≈4.5110\pm 2\pi k\)

Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as \(x\) or \(u\). If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Example \(\PageIndex{6A}\): Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: \({\cos}^2 \theta+3 \cos \theta−1=0\), \(0≤\theta<2\pi\).

We begin by using substitution and replacing \(\cos \theta\) with \(x\). It is not necessary to use substitution, but it may make the problem easier to solve visually. Let \(\cos \theta=x\). We have

\(x^2+3x−1=0\)

The equation cannot be factored, so we will use the quadratic formula: \(x=\dfrac{−b\pm \sqrt{b^2−4ac}}{2a}\).

\[\begin{align*} x&= \dfrac{ -3\pm \sqrt{ {(-3)}^2-4 (1) (-1) } }{2}\\ &= \dfrac{-3\pm \sqrt{13}}{2}\end{align*}\]

Replace \(x\) with \(\cos \theta \) and solve.

\[\begin{align*} \cos \theta&= \dfrac{-3\pm \sqrt{13}}{2}\\ \theta&= {\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right) \end{align*}\]

Note that only the + sign is used. This is because we get an error when we solve \(\theta={\cos}^{−1}\left(\dfrac{−3−\sqrt{13}}{2}\right)\) on a calculator, since the domain of the inverse cosine function is \([ −1,1 ]\). However, there is a second solution:

\[\begin{align*} \theta&= {\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right)\\ &\approx 1.26 \end{align*}\]

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

\[\begin{align*} \theta&= 2\pi-{\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right)\\ &\approx 5.02 \end{align*}\]

Example \(\PageIndex{6B}\): Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: \(2 {\sin}^2 \theta−5 \sin \theta+3=0\), \(0≤\theta≤2\pi\).

Using grouping, this quadratic can be factored. Either make the real substitution, \(\sin \theta=u\),or imagine it, as we factor:

\[\begin{align*} 2 {\sin}^2 \theta-5 \sin \theta+3&= 0\\ (2 \sin \theta-3)(\sin \theta-1)&= 0 \qquad \text {Now set each factor equal to zero.}\\ 2 \sin \theta-3&= 0\\ 2 \sin \theta&= 3\\ \sin \theta&= \dfrac{3}{2}\\ \sin \theta-1&= 0\\ \sin \theta&= 1 \end{align*}\]

Next solve for \(\theta\): \(\sin \theta≠\dfrac{3}{2}\), as the range of the sine function is \([ −1,1 ]\). However, \(\sin \theta=1\), giving the solution \(\theta=\dfrac{\pi}{2}\).

Make sure to check all solutions on the given domain as some factors have no solution.

Exercise \(\PageIndex{4}\)

Solve \({\sin}^2 \theta=2 \cos \theta+2\), \(0≤\theta≤2\pi\). [Hint: Make a substitution to express the equation only in terms of cosine.]

\(\cos \theta=−1\), \(\theta=\pi\)

Example \(\PageIndex{7A}\): Solving a Trigonometric Equation Using Algebra

Solve exactly: \(2 {\sin}^2 \theta+\sin \theta=0;\space 0≤\theta<2\pi\)

This problem should appear familiar as it is similar to a quadratic. Let \(\sin \theta=x\). The equation becomes \(2x^2+x=0\). We begin by factoring:

\[\begin{align*} 2x^2+x&= 0\\ x(2x+1)&= 0\qquad \text {Set each factor equal to zero.}\\ x&= 0\\ 2x+1&= 0\\ x&= -\dfrac{1}{2} \end{align*}\] Then, substitute back into the equation the original expression \(\sin \theta \) for \(x\). Thus, \[\begin{align*} \sin \theta&= 0\\ \theta&= 0,\pi\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\dfrac{11\pi}{6} \end{align*}\]

The solutions within the domain \(0≤\theta<2\pi\) are \(\theta=0,\pi,\dfrac{7\pi}{6},\dfrac{11\pi}{6}\).

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

\[\begin{align*} {\sin}^2 \theta+\sin \theta&= 0\\ \sin \theta(2\sin \theta+1)&= 0\\ \sin \theta&= 0\\ \theta&= 0,\pi\\ 2 \sin \theta+1&= 0\\ 2\sin \theta&= -1\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\dfrac{11\pi}{6} \end{align*}\]

We can see the solutions on the graph in Figure \(\PageIndex{3}\). On the interval \(0≤\theta<2\pi\),the graph crosses the \(x\) - axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

Graph of 2*(sin(theta))^2 + sin(theta) from 0 to 2pi. Zeros are at 0, pi, 7pi/6, and 11pi/6.

We can verify the solutions on the unit circle in via the result in the section on Sum and Difference Identities as well.

Example \(\PageIndex{7B}\): Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: \(2 {\sin}^2 \theta−3 \sin \theta+1=0\), \(0≤\theta<2\pi\).

We can factor using grouping. Solution values of \(\theta\) can be found on the unit circle.

\[\begin{align*} (2 \sin \theta-1)(\sin \theta-1)&= 0\\ 2 \sin \theta-1&= 0\\ \sin \theta&= \dfrac{1}{2}\\ \theta&= \dfrac{\pi}{6}, \dfrac{5\pi}{6}\\ \sin \theta&= 1\\ \theta&= \dfrac{\pi}{2} \end{align*}\]

Exercise \(\PageIndex{5}\)

Solve the quadratic equation \(2{\cos}^2 \theta+\cos \theta=0\).

\(\dfrac{\pi}{2}, \space \dfrac{2\pi}{3}, \space \dfrac{4\pi}{3}, \space \dfrac{3\pi}{2}\)

Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

Example \(\PageIndex{8}\): Solving an Equation Using an Identity

Solve the equation exactly using an identity: \(3 \cos \theta+3=2 {\sin}^2 \theta\), \(0≤\theta<2\pi\).

If we rewrite the right side, we can write the equation in terms of cosine:

\[\begin{align*} 3 \cos \theta+3&= 2 {\sin}^2 \theta\\ 3 \cos \theta+3&= 2(1-{\cos}^2 \theta)\\ 3 \cos \theta+3&= 2-2{\cos}^2 \theta\\ 2 {\cos}^2 \theta+3 \cos \theta+1&= 0\\ (2 \cos \theta+1)(\cos \theta+1)&= 0\\ 2 \cos \theta+1&= 0\\ \cos \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{2\pi}{3},\space \dfrac{4\pi}{3}\\ \cos \theta+1&= 0\\ \cos \theta&= -1\\ \theta&= \pi\\ \end{align*}\]

Our solutions are \(\theta=\dfrac{2\pi}{3},\space \dfrac{4\pi}{3},\space \pi\).

Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as \(\sin(2x)\) or \(\cos(3x)\). When confronted with these equations, recall that \(y=\sin(2x)\) is a horizontal compression by a factor of 2 of the function \(y=\sin x\). On an interval of \(2\pi\),we can graph two periods of \(y=\sin(2x)\),as opposed to one cycle of \(y=\sin x\). This compression of the graph leads us to believe there may be twice as many x -intercepts or solutions to \(\sin(2x)=0\) compared to \(\sin x=0\). This information will help us solve the equation.

Example \(\PageIndex{9}\): Solving a Multiple Angle Trigonometric Equation

Solve exactly: \(\cos(2x)=\dfrac{1}{2}\) on \([ 0,2\pi )\).

We can see that this equation is the standard equation with a multiple of an angle. If \(\cos(\alpha)=\dfrac{1}{2}\),we know \(\alpha\) is in quadrants I and IV. While \(\theta={\cos}^{−1} \dfrac{1}{2}\) will only yield solutions in quadrants I and II, we recognize that the solutions to the equation \(\cos \theta=\dfrac{1}{2}\) will be in quadrants I and IV.

Therefore, the possible angles are \(\theta=\dfrac{\pi}{3}\) and \(\theta=\dfrac{5\pi}{3}\). So, \(2x=\dfrac{\pi}{3}\) or \(2x=\dfrac{5\pi}{3}\), which means that \(x=\dfrac{\pi}{6}\) or \(x=\dfrac{5\pi}{6}\). Does this make sense? Yes, because \(\cos\left(2\left(\dfrac{\pi}{6}\right)\right)=\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}\).

Are there any other possible answers? Let us return to our first step.

In quadrant I, \(2x=\dfrac{\pi}{3}\), so \(x=\dfrac{\pi}{6}\) as noted. Let us revolve around the circle again:

\[\begin{align*} 2x&= \dfrac{\pi}{3}+2\pi\\ &= \dfrac{\pi}{3}+\dfrac{6\pi}{3}\\ &= \dfrac{7\pi}{3}\\ x&= \dfrac{7\pi}{6}\\ \text {One more rotation yields}\\ 2x&= \dfrac{\pi}{3}+4\pi\\ &= \dfrac{\pi}{3}+\dfrac{12\pi}{3}\\ &= \dfrac{13\pi}{3}\\ \end{align*}\]

\(x=\dfrac{13\pi}{6}>2\pi\), so this value for \(x\) is larger than \(2\pi\), so it is not a solution on \([ 0,2\pi )\).

In quadrant IV, \(2x=\dfrac{5\pi}{3}\), so \(x=\dfrac{5\pi}{6}\) as noted. Let us revolve around the circle again:

\[\begin{align*} 2x&= \dfrac{5\pi}{3}+2\pi\\ &= \dfrac{5\pi}{3}+\dfrac{6\pi}{3}\\ &= \dfrac{11\pi}{3} \end{align*}\]

so \(x=\dfrac{11\pi}{6}\).

One more rotation yields

\[\begin{align*} 2x&= \dfrac{5\pi}{3}+4\pi\\ &= \dfrac{5\pi}{3}+\dfrac{12\pi}{3}\\ &= \dfrac{17\pi}{3} \end{align*}\]

\(x=\dfrac{17\pi}{6}>2\pi\),so this value for \(x\) is larger than \(2\pi\),so it is not a solution on \([ 0,2\pi )\)  .

Our solutions are \(x=\dfrac{\pi}{6}, \space \dfrac{5\pi}{6}, \space \dfrac{7\pi}{6}\), and \(\dfrac{11\pi}{6}\). Note that whenever we solve a problem in the form of \(sin(nx)=c\), we must go around the unit circle \(n\) times.

Key Concepts

  • When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution. See Example \(\PageIndex{1}\), Example \(\PageIndex{2}\), and Example \(\PageIndex{3}\).
  • Equations involving a single trigonometric function can be solved or verified using the unit circle. See Example \(\PageIndex{4}\), Example \(\PageIndex{5}\), and Example \(\PageIndex{6}\), and Example \(\PageIndex{7}\).
  • We can also solve trigonometric equations using a graphing calculator. See Example \(\PageIndex{8}\) and Example \(\PageIndex{9}\).
  • Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. See Example \(\PageIndex{10}\), Example \(\PageIndex{11}\), Example \(\PageIndex{12}\), and Example \(\PageIndex{13}\).
  • We can also use the identities to solve trigonometric equation. See Example \(\PageIndex{14}\), Example \(\PageIndex{15}\), and Example \(\PageIndex{16}\).
  • We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. See Example \(\PageIndex{17}\).
  • Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. See Example \(\PageIndex{18}\).

Contributors and Attributions

Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a  Creative Commons Attribution License 4.0  license. Download for free at  https://openstax.org/details/books/precalculus .

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Trigonometric Equations

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  • Sahithi venkatesan

To solve a trigonometric equation, we need the following preliminary knowledge:

If \(\sin \theta = \sin \alpha\), then \(\theta=n\pi+(-1)^{n}\alpha\). Thus, if \(n\) is odd, \(\theta=(2m+1)\pi-\alpha,\) and if \(n\) is even, \(\theta=2m\pi+\alpha\).

If \(\cos \theta = \cos \alpha\), then \(\theta = 2n\pi\pm\alpha\).

If \(\tan \theta = \tan\alpha\), then \(\theta=n\pi+\alpha\).

These hold true for integers \(n,m\).

Now on to solving equations. The general method of solving an equation is to convert it into the form of one ratio only. Then, using these results, we can obtain solutions.

Solving basic equations can be taken care of with the trigonometric R method .

Consider the following example:

Solve the following equation: \[\cos x - \sqrt{3}\sin x = 2.\] Using the R method, this equation can be converted into \[\cos\left(x+\frac{\pi}{3}\right)=1.\] Now, we know that \(\cos0^ \circ=1\). Hence we rewrite the equation as \[\cos\left(x+\frac{\pi}{3}\right)=\cos0^\circ.\] Using the above results, we have \[\begin{align} x+\frac{\pi}{3} &= 2n\pi\pm0\\ x&=2n\pi-\frac{\pi}{3}. \ _\square \end{align}\] Note: Sometimes the question may also provide a range for \(x\). In this case, we take various integral values for \(x\) and find the solutions.

But as the equations get harder, a variety of techniques come handy: the double- and triple-angle formulas, the sum-to-product formulas, etc. Sometimes the equation is converted into a form where the R method can be used, or sometimes we have quadratic equations in the field.

Let's take another example.

Find the general solution of the equation \(\cos2x-2\tan x+2=0.\) Note that \(\cos 2x = \frac{1-\tan^{2} x}{1+\tan^{2} x}.\) Substituting this, the equation has only one variable: \(\tan x\). We evaluate and then factorize the expression to obtain \[(\tan x - 1)\left(2\tan^{2} x + \tan x + 3\right)=0.\] As the equation \(2\tan^{2} x + \tan x + 3 = 0\) has imaginary roots, we do not consider it. Hence, we have \[\begin{align} \tan x - 1 &= 0\\ \tan x &= 1\\ x&=\frac{\pi}{4}+n\pi. \ _\square \end{align}\]

Specific Solutions - Basic

Specific solutions - intermediate, general solutions - basic, general solutions - intermediate, factoring - basic, factoring - intermediate, problem solving - basic, problem solving - intermediate.

What are the solutions of the equation \[ \sin 2x = \frac{\sqrt{3}}{2} \] in the \(x\)-interval \( [0, 2\pi] ?\) Since \( 0 \leq 2x \leq 4\pi ,\) we have \[\begin{align} 2x&= \frac{\pi}{3}, \frac{2}{3} \pi, \frac{7}{3} \pi, \frac{8}{3} \pi\\\\ \Rightarrow x&= \frac{\pi}{6}, \frac{\pi}{3}, \frac{7}{6} \pi, \frac{4}{3} \pi. \ _\square \end{align} \]
What are the solutions of \[ \sin \left(x-\frac{\pi}{3} \right) = -\frac{1}{2} \] in the \(x\)-interval \( [0, 2\pi] ?\) Since \( -\frac{\pi}{3} \leq x-\frac{\pi}{3} \leq \frac{5}{3} \pi ,\) we have \[\begin{align} x-\frac{\pi}{3} &=-\frac{\pi}{6}, \frac{7}{6} \pi\\\\ \Rightarrow x &= \frac{\pi}{6}, \frac{3}{2} \pi. \ _\square \end{align}\]
What is the general solution of \[ \tan x = 1 ?\] In the first period \( 0 \leq x < \pi,\) the solution for this equation is \(x= \frac{\pi}{4}.\) Since the period of \(\tan x\) is \(\pi,\) the general solution of the given equation is \[ x = n \pi + \frac{\pi}{4} . \ _\square\]
What is the general solution of \[ \cos 2x +3\cos x -1=0 ?\] Since \( \cos 2x = 2\cos^2 x -1,\) we have \[ \begin{align} 2 \cos^2 x -1 + 3\cos x -1 &= 0 \\ (2\cos x -1)(\cos x +2) &= 0\\ \cos x &= \frac{1}{2}. \qquad (\text{since }\lvert \cos x \rvert \leq 1) \end{align} \] Since the period of \( \cos x \) is \( 2\pi,\) the general solution for the given equation is \[ x = 2n \pi \pm \frac{\pi}{3}. \ _\square \]
What is the general solution of \[ \tan 2x - 3\tan x = 0 ?\] Since \( \tan 2x = \frac{2\tan x}{1-\tan^2 x} ,\) we have \[ \begin{align} \tan 2x - 3\tan x &= 0 \\ \frac{2\tan x}{1-\tan^2 x} -3\tan x &= 0 \\ 2\tan x -3\tan x \left(1-\tan^2 x\right) &= 0 \\ \tan x\left(3\tan^2 x -1\right) &= 0 \\ \Rightarrow \tan x &= 0, \pm \frac{1}{\sqrt{3}}. \end{align} \] Since the period of \( \tan x\) is \(\pi,\) the general solution for the given equation is \[ x= n \pi, \text{ or } x=n \pi \pm \frac{\pi}{6}. \ _\square\]
What are the solutions of the following equation for \(0 \leq x \leq 2\pi:\) \[ 2\cos^2 x - \sin x -1=0 ? \] Since \( \sin^2 + \cos^2 = 1 ,\) we rewrite the given equation to obtain \[ \begin{align} 2 \cos^2 x-\sin x -1 &= 0 \\ \Rightarrow 2(1-\sin^2 x)-\sin x -1 &= 0 \\ \left(2\sin x - 1\right)(\sin x + 1) &=0 \\ \sin x &= \frac{1}{2}, -1. \end{align} \] Since \(0 \leq x \leq 2\pi,\) for \( \sin x = \frac{1}{2}\) we have \[ x = \frac{\pi}{6}, \frac{5}{6} \pi. \qquad (1) \] For \( \sin x = -1,\) we have \[ x = \frac{3}{2} \pi. \qquad (2) \] Thus, from \( (1)\) and \( (2)\) we have \[ x = \frac{\pi}{6}, \frac{5}{6} \pi, \frac{3}{2} \pi. \ _\square\]
What are the solutions of the following equation for \(0 \leq x \leq 2\pi:\) \[ \sin^ 4 x + \sin^2 x -1 = \cos^4 x + \cos^2 x ?\] Using \(\sin^2 x + \cos^2 x = 1,\) we have \[ \begin{align} \sin^ 4 x + \sin^2 x -1 &= \cos^4 x + \cos^2 x \\ \sin^4 x - \cos^4 x + \sin^2 x - \cos^2 x -1 &= 0 \\ \left(\sin^2 x + \cos^2 x\right)\left(\sin^2 x - \cos^2 x\right) + \sin^2x - \cos^2 x -1 &= 0 \\ 2\sin^2 x - 2\cos^2 x - 1 &= 0 \\ 2\sin^2 x - 2\left(1-\sin^2 x\right) -1 &= 0 \\ \Rightarrow \sin^2 x &= \frac{3}{4} \\ \sin x &= \pm \frac{\sqrt{3}}{2}. \end{align} \] Since \(0 \leq x \leq 2\pi,\) for \( \sin x = \frac{\sqrt{3}}{2}\) we have \[ x = \frac{\pi}{3}, \frac{2}{3} \pi . \qquad (1) \] For \( \sin x = -\frac{\sqrt{3}}{2} ,\) we have \[ x= \frac{4}{3} \pi, \frac{5}{3} \pi. \qquad (2) \] Thus, from \( (1) \) and \( (2) \) the solutions are \[x = \frac{\pi}{3}, \frac{2}{3} \pi, \frac{4}{3} \pi, \frac{5}{3} \pi. \ _\square\]
What are the solutions of the following equation for \(0 \leq x \leq 2\pi:\) \[ \cos^2 x - \sin^2 2x = 0?\] We have \[ \begin{align} \cos^2 x - \sin^2 2x &= 0 \\ \cos^2 x - 4\sin^2 x \cos^2 x &= 0 \qquad ( \text{ since } \sin 2x = 2\sin x \cos x ) \\ \cos^2x(1-4\sin^2 x) &= 0 \\ \Rightarrow \cos x &= 0, ~\sin x = \frac{1}{2}, - \frac{1}{2}. \end{align} \] Since \(0 \leq x \leq 2\pi,\) for \( \cos x =0\) we have \[ x = \frac{\pi}{2}, \frac{3}{2} \pi. \qquad (1) \] For \( \sin x = \frac{1}{2},\) we have \[ x = \frac{\pi}{6}, \frac{5}{6} \pi. \qquad (2) \] For \( \sin x = -\frac{1}{2},\) we have \[ x= \frac{7}{6} \pi, \frac{11}{6} \pi. \qquad (3) \] Thus, from \( (1), (2) \) and \((3)\) the solutions are \[ x= \frac{\pi}{2}, \frac{3}{2} \pi, \frac{\pi}{6}, \frac{5}{6} \pi, \frac{7}{6} \pi, \frac{11}{6} \pi. \ _\square\]
What are the values of \(\alpha\) in the interval \( [0, 2\pi]\) such that the quadratic equation \( x^2 + 2x + 2\cos \alpha=0\) has repeated roots? For the quadratic equation \( x^2 + 2x +2\cos \alpha = 0\) to have repeated roots, its discriminant \(D\) must be 0. Thus, \[ \begin{align} \frac{D}{4} = 1-2\cos \alpha &= 0 \\ \cos \alpha &= \frac{1}{2} \\ \Rightarrow \alpha &= \frac{\pi}{3}, \frac{5}{3} \pi. \ _\square \end{align} \]

\[\sin\theta+\cos\theta=\sqrt{2}\sin(90^{\circ}-\theta), \quad \cot\theta = \, ? \]

Give your answer to the above problem to 3 decimal places.

If a root of the equation \[1-6\sin x = \sqrt{5-12\sin x} \] is \( \alpha,\) what is the value of \( \sin \alpha \cos 2\alpha?\) Squaring both sides, we have \[ \begin{align} (1-6\sin x)^2 &= \left(\sqrt{5-12\sin x}\right)^2 \\ 1-12\sin x + 36\sin^2 x &= 5 - 12 \sin x \\ \sin^2 x &= \frac{1}{9} \\ \Rightarrow \sin x &= \pm \frac{1}{3}. \end{align} \] Since only \(\sin x = - \frac{1}{3} \) satisfies \( 1-6\sin x = \sqrt{5-12\sin x} ,\) we have \( \sin \alpha= -\frac{1}{3}.\) Thus, \( \sin \alpha \cos 2\alpha \) is \[ \begin{align} \sin \alpha \cos 2\alpha &= \sin \alpha\left(1-2\sin^2 \alpha\right) \\ &= -\frac{1}{3}\left( 1- \frac{2}{9} \right ) \\ &= -\frac{7}{27}. \ _\square \end{align} \]
If the roots of the quadratic equation \( 8x^2+7x+a=0\) are \( \sin \theta\) and \( \cos 2\theta,\) what is \(a?\) From Vieta's formulas , we have \[ \begin{align} \sin \theta + \cos 2\theta &= -\frac{7}{8} &\qquad (1) \\ \sin \theta \cos 2\theta &= \frac{a}{8}. &\qquad (2) \end{align} \] From \((1),\) we have \[ \begin{align} \sin \theta + \cos 2\theta &= -\frac{7}{8} \\ 8(\sin \theta +1 - 2 \sin^2 \theta) + 7 &= 0 \\ (4\sin \theta +3)(4\sin \theta - 5) &= 0 \\ \Rightarrow \sin \theta &= -\frac{3}{4}, \frac{5}{4}. \end{align} \] Since \(\lvert \sin \theta \rvert \leq 1,\) it follows that \( \sin \theta = -\frac{3}{4}.\) Thus, we have \[ \begin{align} \cos 2\theta &= 1-2\sin^2 \theta \\ &= 1-2\sin^2\left(-\frac{3}{4} \right) \\ &= -\frac{1}{8}. \end{align} \] Substituting this into \((2)\) gives \[ \begin{align} \sin \theta \cos 2 \theta &= \frac{a}{8} \\ \left( -\frac{3}{4} \right) \left( -\frac{1}{8} \right ) &= \frac{a}{8} \\ \Rightarrow a&= \frac{3}{4}. \ _\square \end{align} \]

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Solving Triangles

"Solving" means finding missing sides and angles.

Six Different Types

If you need to solve a triangle right now choose one of the six options below:

Which Sides or Angles do you know already? (Click on the image or link)

... or read on to find out how you can become an expert triangle solver :

Your Solving Toolbox

Want to learn to solve triangles?

Imagine you are " The Solver " ... ... the one they ask for when a triangle needs solving!

In your solving toolbox (along with your pen, paper and calculator) you have these 3 equations:

1. Angles Add to 180° :

A + B + C = 180°

When you know two angles you can find the third.

2. Law of Sines (the Sine Rule):

When there is an angle opposite a side, this equation comes to the rescue.

Note: angle A is opposite side a, B is opposite b, and C is opposite c.

3. Law of Cosines (the Cosine Rule):

This is the hardest to use (and remember) but it is sometimes needed to get you out of difficult situations.

It is an enhanced version of the Pythagoras Theorem that works on any triangle.

Six Different Types (More Detail)

There are SIX different types of puzzles you may need to solve. Get familiar with them:

This means we are given all three angles of a triangle, but no sides.

AAA triangles are impossible to solve further since there are is nothing to show us size ... we know the shape but not how big it is.

We need to know at least one side to go further. See Solving "AAA" Triangles .

This mean we are given two angles of a triangle and one side, which is not the side adjacent to the two given angles.

Such a triangle can be solved by using Angles of a Triangle to find the other angle, and The Law of Sines to find each of the other two sides. See Solving "AAS" Triangles .

This means we are given two angles of a triangle and one side, which is the side adjacent to the two given angles.

In this case we find the third angle by using Angles of a Triangle , then use The Law of Sines to find each of the other two sides. See Solving "ASA" Triangles .

This means we are given two sides and the included angle.

For this type of triangle, we must use The Law of Cosines first to calculate the third side of the triangle; then we can use The Law of Sines to find one of the other two angles, and finally use Angles of a Triangle to find the last angle. See Solving "SAS" Triangles .

This means we are given two sides and one angle that is not the included angle.

In this case, use The Law of Sines first to find either one of the other two angles, then use Angles of a Triangle to find the third angle, then The Law of Sines again to find the final side. See Solving "SSA" Triangles .

This means we are given all three sides of a triangle, but no angles.

In this case, we have no choice. We must use The Law of Cosines first to find any one of the three angles, then we can use The Law of Sines (or use The Law of Cosines again) to find a second angle, and finally Angles of a Triangle to find the third angle. See Solving "SSS" Triangles .

Tips to Solving

Here is some simple advice:

When the triangle has a right angle, then use it, that is usually much simpler.

When two angles are known, work out the third using Angles of a Triangle Add to 180° .

Try The Law of Sines before the The Law of Cosines as it is easier to use.

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Sat / act prep online guides and tips, the 36 trig identities you need to know.

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General Education

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If you’re taking a geometry or trigonometry class, one of the topics you’ll study are trigonometric identities. There are numerous trig identities, some of which are key for you to know, and others that you’ll use rarely or never. This guide explains the trig identities you should have memorized as well as others you should be aware of. We also explain what trig identities are and how you can verify trig identities.

In math, an "identity" is an equation that is always true, every single time. Trig identities are trigonometry equations that are always true, and they’re often used to solve trigonometry and geometry problems and understand various mathematical properties. Knowing key trig identities helps you remember and understand important mathematical principles and solve numerous math problems.

The 25 Most Important Trig Identities

Below are six categories of trig identities that you’ll be seeing often. Each of these is a key trig identity and should be memorized. It seems like a lot at first, but once you start studying them you’ll see that many follow patterns that make them easier to remember.

Basic Identities

These identities define the six trig functions.

$$sin(θ) = 1/{csc(θ)}$$

$$cos(θ) = 1/{sec(θ)}$$

$$tan(θ) = 1/{cot(θ)} = {sin(θ)}/{cos(θ)}$$

$$csc(θ) = 1/{sin(θ)}$$

$$sec(θ) = 1/{cos(θ)}$$

$$cot(θ) = 1/{tan(θ)} = {cos(θ)}/{sin(θ)}$$

Pythagorean Identities

These identities are the trigonometric proof of the Pythagorean theorem (that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides, or $a^2 + b^2 = c^2$). The first equation below is the most important one to know, and you’ll see it often when using trig identities.

$$sin^2(θ) + cos^2(θ) = 1$$

$$tan^2(θ) + 1 = sec^2(θ)$$

$$1 + cot^2(θ) = csc^2(θ)$$

Co-function Identities

Each of the trig functions equals its co-function evaluated at the complementary angle.

$$sin(θ) = cos({π/2} - θ)$$

$$cos(θ) = sin({π/2} - θ)$$

$$tan(θ) = cot({π/2} - θ)$$

$$cot(θ) = tan({π/2} - θ)$$

$$csc(θ) = sec({π/2} - θ)$$

$$sec(θ) = csc({π/2} - θ)$$

Negative Angle Identities

Sine, tangent, cotangent, and cosecant are odd functions (symmetric about the origin). Cosine and secant are even functions (symmetric about the y-axis).

$$sin(-θ) = -sin(θ)$$

$$cos(-θ) = cos(θ)$$

$$tan(-θ) = -tan(θ)$$

Sum and Difference Identities

These are sometimes known as Ptolemy’s Identities as he’s the one who first proved them.

$$sin(α + β) = sin(α)cos(β) + cos(α)sin(β)$$

$$sin(α – β) = sin(α)cos(β) – cos(α)sin(β)$$

$$cos(α + β) = cos(α)cos(β) – sin(α)sin(β)$$

$$cos(α – β) = cos(α)cos(β) + sin(α)sin(β)$$

Double-Angle Identities

You only need to memorize one of the double-angle identities for cosine. The other two can be derived from the Pythagorean theorem by using the identity $sin^2(θ) + cos^2(θ) = 1$ to convert one cosine identity to the others.

$$sin(2θ) = 2 sin(θ) cos(θ)$$

$$cos(2θ) = cos^2(θ) – sin^2(θ) = 1 – 2 sin^2(θ) = 2 cos^2(θ) – 1$$

$$tan(2θ)={2 tan(θ)}/{1– tan^2(θ)}$$

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Additional Trig Identities

These three categories of trig identities are used less often. You should look through them to make sure you understand them, but they typically don’t need to be memorized.

Half-Angle Identities

These are inversions of the double-angle identities.

$$sin2(θ) = {1/2}(1-cos (2θ))$$

$$cos2(θ) = {1/2}(1+ cos (2θ))$$

$$tan2(θ) = {1-cos(2θ)}/{1+ cos (2θ)}$$

Sum Identities

These trig identities make it possible for you to change a sum or difference of sines or cosines into a product of sines and cosines.

$$sin(α) + sin(β)= 2sin({α + β}/ 2)  cos({α - β}/ 2)$$

$$sin(α) - sin(β)= 2cos({α + β}/ 2)  sin({α - β}/ 2)$$

$$cos(α) + cos(β)= 2cos({α + β} / 2)  cos({α - β}/ 2)$$

$$cos(α) - cos(β)= -2sin ({α + β}/ 2)  sin({α - β}/ 2)$$

Product Identities

This group of trig identities allows you to change a product of sines or cosines into a product or difference of sines and cosines.

$$sin(α) cos(β)= {1/2}(sin (α + β) + sin (α - β))$$

$$cos(α) sin(β)= {1/2}(sin (α + β) - sin (α - β))$$

$$sin(α) sin(β)= {1/2}(cos (α - β) - cos(α + β))$$

$$cos(α) cos(β)= {1/2}(cos (α - β) + cos(α + β))$$

Verifying Trigonometric Identities

Once you have gone over all the key trig identities in your math class, the next step will be verifying them. Verifying trig identities means making two sides of a given equation identical to each other in order to prove that it is true. You’ll use trig identities to alter one or both sides of the equation until they’re the same.

Verifying trig identities can require lots of different math techniques, including FOIL, distribution, substitutions, and conjugations. Each equation will require different techniques, but there are a few tips to keep in mind when verifying trigonometric identities.

#1: Start With the Harder Side

Despite what you may initially want to do, we recommend starting with the side of the equation that looks messier or more difficult.  Complicated-looking equations often give you more possibilities to try out than simpler equations, so start with the trickier side so you have more options.

#2: Remember That You Can Change Both Sides

You don’t need to stick to only changing one side of the equation. If you get stuck on one side, you can switch over to the other side and begin changing it as well. Neither side of the equation needs to be the same as how it was originally; as long as both sides of the equation end up being identical, the identity has been verified.

#3: Turn all the Functions Into Sines and Cosines

Most students learning trig identities feel most comfortable with sines and cosines because those are the trig functions they see the most. Make things easier on yourself by converting all the functions to sines and cosines!

body_mathequations

Verify the identity $cos(θ)sec(θ) = 1$

Let’s change that secant to a cosine. Using basic identities, we know $sec(θ) = 1/{cos(θ)}$. That gives us:

$$cos(θ) (1/{cos(θ)}) = 1$$

The cosines on the left cancel each other out, leaving us with $1=1$.

Identity verified!

Verify the identity $1 − cos(2θ) = tan(θ) sin(2θ)$

Let’s start with the left side since it has more going on. Using basic trig identities, we know tan(θ) can be converted to sin(θ)/ cos(θ), which makes everything sines and cosines.

$$1 − cos(2θ) = ({sin(θ)}/{cos(θ)}) sin(2θ)$$

Distribute the right side of the equation:

$$1 − cos(2θ) = 2sin^2(θ)$$

There are no more obvious steps we can take to transform the right side of the equation, so let’s move to the left side. We can use the Pythagorean identity to convert $cos(2θ)$ to $1 - 2sin^2(θ)$

$$1 - (1 - 2sin^2(θ)) = 2sin^2(θ)$$

Now work out the left side of the equation

$$2sin^2(θ) = 2sin^2(θ)$$

The two sides are identical, so the identity has been verified!

Verify the identity $sec(-θ) = sec(θ)$

The left side of the equation is a bit more complicated, so let’s change that secant into a sine or cosine. From the basic trig identities, we know that $sec(θ) = 1/{cos(θ)}$, which means that $sec(-θ) = 1/{cos(-θ)}$. Substitute that for the left side:

$$1/{cos(-θ)} = sec(θ)$$

The negative angle identities tell us that $cos(-θ) = cos(θ)$, so sub that:

$$1/{cos(θ)} = sec(θ)$$

Again, we know that $sec(θ) = 1/{cos(θ)}$, so we end up with:

$$sec(θ) = sec(θ)$$

Summary: Trig Identities Solver

You’ll need to have key trig identities memorized in order to do well in your geometry or trigonometry classes. While there may seem to be a lot of trigonometric identities, many follow a similar pattern, and not all need to be memorized.

When verifying trig identities, keep the following three tips in mind:

  • Start with the trickier side
  • Remember that you can change both sides of the equation
  • Turn the functions into sines and cosines

What's Next?

Wondering which math classes to take in high school? Learn the best math classes for high school students to take by reading our guide!

Wondering whether you should take AB or BC Calculus ? Our guide lays out the differences between the two classes  and explains who should take each course.

Interested in math competitions like the International Math Olympiad ? See our guide for passing the qualifying tests.

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Christine graduated from Michigan State University with degrees in Environmental Biology and Geography and received her Master's from Duke University. In high school she scored in the 99th percentile on the SAT and was named a National Merit Finalist. She has taught English and biology in several countries.

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How to Solve Trig Identities and Tips on Proving Trigonometric Identities

How to Solve Trig Identities and Tips on Proving Trigonometric Identities

Essential Identities

Lean to solve trig identities

How to Solve Trig Identities

The following seven step process will work every time. It is rather tedious, and can take more time than necessary. As you gain more practice, you can skip or combine these steps when you recognize other identities. STEP 1: Convert all sec, csc, cot, and tan to sin and cos. Most of this can be done using the quotient and reciprocal identities. STEP 2: Check all the angles for sums and differences and use the appropriate identities to remove them. STEP 3: Check for angle multiples and remove them using the appropriate formulas. STEP 4: Expand any equations you can, combine like terms, and simplify the equations. STEP 5: Replace cos powers greater than 2 with sin powers using the Pythagorean identities. STEP 6: Factor numerators and denominators, then cancel any common factors. STEP 7: Now, both sides should be exactly equal, or obviously equal, and you have proven your identity.

Example Problem Using the 7 Step Method

Show that cos4(x) - sin4(x) = cos(2x) STEP 1: Everything is already in sin and cos, so this part is done. cos4(x) - sin4(x) = cos (2x) STEP 2: Since there are no sums or difference inside the angles, this part is done. cos4(x) - sin4(x) = cos (2x) STEP 3: cos(2x) is a double angle. Use the double angle formula: cos (2x) = cos2(x) - sin2(x), to simplify. cos4(x) - sin4(x) = cos2(x) - sin2(x) STEP 4: Here is where your algebra knowledge comes in. In this case, we can see that the left side is a “difference of two squares”

[if you forgot: a2-b2 = (a+b)(a-b)]

Left side: cos4x - sin4x - (cos2(x))2 - (cos2(x))2 = (cos2(x)-sin2(x))(cos2(x)+sin2(x)) Now, our problem looks like this: (cos2(x)-sin2x))(cos2(x)+sin2(x))= cos2(x) - sin2(x) The sides are almost the same STEP 5: There are no powers greater than 2, so we can skip this step STEP 6: Since cos2(x) - sin2(x) appears on both sides of the equation, we can cancel it. We are left with: cos2(x) + sin2(x) = 1 STEP 7: Since this is one of the Pythagorean identities, we know it is true, and the problem is done.

Proving Trigonometric Identities

The 7 step method works both sides and meets in the middle, like a V. Some teachers will ask you to prove the identity directly (from one side to the other in a straight line). That is easily done using the work above. Just write down all the left side parts in order first, then the right side parts in backwards order, so it looks like this: cos4(x) - sin4(x) = (cos2(x)-sin2(x))(cos2(x)+sin2(x)) = (cos2(x)-sin2(x))(1) = cos2(x)-sin2x = cos(2x) Or write the right hand steps in order first and then the left hand step backwards so it looks like this: cos(2x) = cos2(x)-sin2(x) = (cos2(x)-sin2(x))(cos2(x)+sin2(x) = cos4(x) - sin4(x) Even though this is a simple problem, the same steps will work every time to solve trig identities no matter the difficulty.

Get both sides of the equation in the same functions. You don’t always have to use sin and cos, but its easier to compare when both sides are composed of similar functions

Make sure all your angles are the same. Using sin(2A) and sinA is difficult, but if you use sin2A = 2sin(x)cos(x), that leaves sin(x) and cos(x), and now all your functions match. The same goes for addition and subtraction: don’t try working with sin(A+B) and sinA. Instead, use sin(A+B) = sin(x)cos(x)+cos(x)sin(x) so that all the angles match.

3 main ways to solve: Convert right side to left side [direct right-left], convert left side to right side [direct left-right], or convert both sides to the same function [meet in the middle]

If you need to add more powers (or remove them), use cos^2(x) + sin^2(x) = 1. You can always multiply by 1 without changing the meaning, so therefore you can always multiply by cos^2(x)+sin^2(x).

Once you get the hang of it, you will begin to see patterns. For instance, in the example above, you might notice right off that the left side is difference of two squares and do that first. Then, you quickly simplify to cos^2(x) = sin^2(x), which tells you which double angle identity to use.

If you keep getting stuck on a problem, take a break. Come back with a clean sheet of paper, and start over from the beginning. Often, it helps to change the direction (from left-right to right-left).

Let someone else read through your work, just to see if they follow it and can give a new perspective. When you stare at the same equations for too long, you’ll likely start to miss things that you would have noticed at the beginning.

Continue reading below for additional help on how to solve trig identities! Image by  준원 서  from  Pixabay

This post is part of the series: Trig Help

Everything you need to know to get through your trig class. Whether in high school or college, the tips, tricks, formulas and methods you need can all be found here.

  • A Guide to Proving Trig Identities
  • Ace the Trig Exam With This Study Guide
  • Functions List for Trigonometry

How to Solve Trigonometry Problems

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Introduction: How to Solve Trigonometry Problems

How to Solve Trigonometry Problems

Introduction: Trigonometry. This Instructable is originally intended for the ninth students at DIS, but anybody is welcome to learn about Trigonometry. In this Introduction, I will give a general overview of the topic of Trigonometry, tips on how to learn and study well, and then go into more detail. In Mathematics, it is always important to learn how to understand what you are doing, and why you are doing these steps as opposed to just memorizing it. Trigonometry is the study of triangles. In this instructable, I will start basic with naming the sides of the right triangles, the trigonometric functions, and then gradually increase the difficulty so that the reader can eventually see how to tackle these problems, and apply them to real world situations. I will also provide tips on how to study and learn this topic well. This tutorial requires you to already know some basic algebra and geometry.

Step 1: Formulas and Definitions

Formulas and Definitions

First Slide: Formulas - Sin = Opp/Hyp, Cos = Adj/Hypotenuse Tan = Opp/Adj

Note: x is the angle we are using to determine the opp, adj, or the hypotenuse. If it were another angle, then the opposite and adjacent would change. Tricks : Soh Cah Toa. You may have seen or heard of this many times. The S in Soh represents the Sine, while the o stands for opposite, and the h stands for hypotenuse.

Definitions: Hypotenuse - The longest side of a right triangle. Opposite to the 90° angle.

Opposite - The side opposite to the angle of reference.

Adjacent - The side that is next to the angle of reference that is not the hypotenuse.

Right Triangle - A triangle with one ninety degree angle.

Step 2: Practice Problems

Practice Problems

Second Slide: Steps

a) Please Identify the following sides of the triangles with the appropriate names involving opposite, adjacent, or the hypotenuse.

b) find the sin, cos, tan ratios of the given angle.

c) Solve for side x. (only for the top triangle)

d) use a calculator to find the numerical value of x. (top Triangle)

Tip: Use Pythagoras Theorem To solve for the third unknown side. Opp^2+Adj^2=Hyp^2. Then use algebra to solve for one of these sides.

Answers: Left Triangle- A) Hyp=5m, Adj=4m, Opp=3m B) SinC = ⅗, CosC = ⅘ TanC = ¾

Right Triangle- A) Hyp=x, Adj=unknown side, Opp=2,500.

B) Sin 23 = 2500/x, Cos 23 = unknown side/x, Tan 23 = 2500/unknown side.

C) 1. Sin 23 = 2500/x 2. x Sin 23 = 2500 3. x = 2500/Sin 23. d) Solve with a calculator. Do the same with cos and tan.

Step 3: Finding the Sin and Cos of a Specific Angle

Finding the Sin and Cos of a Specific Angle

Third Slide: Tricks on how to find the value of the sin, cos, tan of a specific angle.

Exp. Sin 30° = 1 (opposite)/2 (Hypotenuse) so it equals ½ = .5(calculator).

Cos 45° = 1/root 2 = .7071 (Calculator). You can use the pythagorean theorem to check that these are valid right triangles.

There are other examples of finding the ratio defining the trigonometric functions of specific angles. The first step is to find the values of the sides, and then divide them. For most angles, however, you will need a calculator. This step was made to help you understand what the strange numbers and decimals on your calculator mean whenever you find the sin, cos, or tan of an angle.

Step 4: Word Problems

Word Problems

Fourth Slide: These are world problems that are found in real-life situations so that you can put your knowledge into more practical use!

1)You have to first identify the right triangle in this scenario.

2)Then identify the parts ie. adj, hyp, and opp.

3)Find the angle you need to use for your situation. What function will give you the side you need to solve for?

4) Apply the function to that angle, solve for the side, and calculate.

Answer: The angle opposite to the 32° angle is also 32°. Use the tan since the adj is given, and the opposite needs to be found. Tan 32° = ?/325, ? = 325 Tan 32°. The crater is 214.827m deep.

Step 5: Inverse Trigonomic Functions

Inverse Trigonomic Functions

Fifth Slide: Inverse trigonometric functions.

The goal is to find the measure of an angle given at least two sides. First, you determine the right function to use (tan, sin, and cos) based off of which sides are given (Hyp, Adj, Opp). Then solve for the angle. Exp. Find X. The first step would be to figure out what is given. The opposite (7) and the hypotenuse (25) are known. What trigonometric function involves both the opposite and the hypotenuse? The sine of course! So we create an equation sinx = 7/25. x = arcsin(7/25). Then just type that into your calculator to find the result. The arcsine is just another word for the inverse sin.

Step 6: What We Have Learned

What We Have Learned

Sixth Slide: Summing it all up. Becoming a better math student.

We have learned what is a right triangle, opp, adj, hyp, sin, cos, tan, how to solve for an unknown side using trigonometry, the pythagorean theorem, values of trigonometric functions for specific angles, applying trigonometry to real world problems, and using the inverse sine to find the value of an angle given the sides. In order to improve, you must practice more math problems. I recommend buying a math book as a source to find a variety of problems, and learn concepts. If you identify your difficulties, be sure to ask for help!

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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.

best way to solve trig problems

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

1

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

best way to solve trig problems

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

2

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

best way to solve trig problems

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

3

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

best way to solve trig problems

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?

best way to solve trig problems

This triangle cannot exist.

best way to solve trig problems

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

best way to solve trig problems

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. 

Screen shot 2020 07 13 at 12.54.08 pm

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:

best way to solve trig problems

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

best way to solve trig problems

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

best way to solve trig problems

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. 

Screen shot 2020 07 13 at 1.38.59 pm

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.

best way to solve trig problems

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.

Screen shot 2020 07 13 at 3.07.05 pm

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:

best way to solve trig problems

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.

Screen shot 2020 07 13 at 5.37.06 pm

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:

best way to solve trig problems

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. 

Screen shot 2020 07 13 at 5.56.45 pm

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?

Screen shot 2020 07 13 at 5.58.09 pm

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  • \sin (x)+\sin (\frac{x}{2})=0,\:0\le \:x\le \:2\pi
  • \cos (x)-\sin (x)=0
  • \sin (4\theta)-\frac{\sqrt{3}}{2}=0,\:\forall 0\le\theta<2\pi
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  • What is tangent equal to?
  • The tangent function (tan), is a trigonometric function that relates the ratio of the length of the side opposite a given angle in a right-angled triangle to the length of the side adjacent to that angle.
  • How to solve trigonometric equations step-by-step?
  • To solve a trigonometric simplify the equation using trigonometric identities. Then, write the equation in a standard form, and isolate the variable using algebraic manipulation to solve for the variable. Use inverse trigonometric functions to find the solutions, and check for extraneous solutions.
  • What is a basic trigonometric equation?
  • A basic trigonometric equation has the form sin(x)=a, cos(x)=a, tan(x)=a, cot(x)=a
  • How to convert radians to degrees?
  • The formula to convert radians to degrees: degrees = radians * 180 / π
  • What is cotangent equal to?
  • The cotangent function (cot(x)), is the reciprocal of the tangent function.cot(x) = cos(x) / sin(x)

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  • Spinning The Unit Circle (Evaluating Trig Functions ) If you’ve ever taken a ferris wheel ride then you know about periodic motion, you go up and down over and over... Read More

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Word Problem: Rachel has 17 apples. She gives some to Sarah. Sarah now has 8 apples. How many apples did Rachel give her?

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Word Problem: Rhonda has 12 marbles more than Douglas. Douglas has 6 marbles more than Bertha. Rhonda has twice as many marbles as Bertha has. How many marbles does Douglas have?

Variables: Rhonda's marbles is represented by (r), Douglas' marbles is represented by (d) and Bertha's marbles is represented by (b)

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Word Problem: if there are 40 cookies all together and Angela takes 10 and Brett takes 5 how many are left?

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Until recently most incumbent industrial companies didn’t use highly advanced software in their products. But now the sector’s leaders have begun applying generative AI and machine learning to all kinds of data—including text, 3D images, video, and sound—to create complex, innovative designs and solve customer problems with unprecedented speed.

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Until recently most incumbent industrial companies didn’t use the most advanced software in their products. But competitors that can extract complex designs, insights, and trends using generative AI have emerged to challenge them.

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  • Vijay Govindarajan is the Coxe Distinguished Professor at Dartmouth College’s Tuck School of Business, an executive fellow at Harvard Business School, and faculty partner at the Silicon Valley incubator Mach 49. He is a New York Times and Wall Street Journal bestselling author. His latest book is Fusion Strategy: How Real-Time Data and AI Will Power the Industrial Future . His Harvard Business Review articles “ Engineering Reverse Innovations ” and “ Stop the Innovation Wars ” won McKinsey Awards for best article published in HBR. His HBR articles “ How GE Is Disrupting Itself ” and “ The CEO’s Role in Business Model Reinvention ” are HBR all-time top-50 bestsellers. Follow him on LinkedIn . vgovindarajan
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COMMENTS

  1. 11 Tips to Conquer Trigonometry Proving

    Tip 1) Always Start from the More Complex Side To prove a trigonometric identity, we always start from either the left hand side (LHS) or the right hand side (RHS) and apply the identities step by step until we reach the other side. However, smart students always start from the more complex side.

  2. Mathway

    Free math problem solver answers your trigonometry homework questions with step-by-step explanations. Mathway. Visit Mathway on the web. Start 7-day free trial on the app. ... While we cover a very wide range of problems, we are currently unable to assist with this specific problem. I spoke with my team and we will make note of this for future ...

  3. Trigonometry

    Unit 1: Right triangles & trigonometry Ratios in right triangles The reciprocal trigonometric ratios Unit 2: Trigonometric functions Unit circle introduction Radians The Pythagorean identity Special trigonometric values in the first quadrant Trigonometric values on the unit circle

  4. How to Solve Trigonometric Equations: A Simple Tutorial

    1 Know the Solving concept. [1] To solve a trig equation, transform it into one or many basic trig equations. Solving trig equations finally results in solving 4 types of basic trig equations. 2 Know how to solve basic trig equations. [2] There are 4 types of basic trig equations: sin x = a ; cos x = a tan x = a ; cot x = a

  5. Introduction to Trigonometry

    Trigonometry can find that missing angle and distance. Or maybe we have a distance and angle and need to "plot the dot" along and up: Questions like these are common in engineering, computer animation and more. And trigonometry gives the answers! Sine, Cosine and Tangent The main functions in trigonometry are Sine, Cosine and Tangent

  6. Trigonometry

    Familiar Attempted Not started Quiz Unit test About this unit Knowing trig identities is one thing, but being able to prove them takes us to another level. In this unit, we'll prove various trigonometric identities and define inverse trigonometric functions, which allow us to solve trigonometric equations.

  7. Right triangles & trigonometry

    Trigonometry 4 units · 36 skills. Unit 1 Right triangles & trigonometry. Unit 2 Trigonometric functions. Unit 3 Non-right triangles & trigonometry. Unit 4 Trigonometric equations and identities. Course challenge. Test your knowledge of the skills in this course. Start Course challenge. Math.

  8. Solving Harder Trig Equations

    The first solution is 45° more than a multiple of 180°, so (180n)° + 45° should do. The second solution is 30° more than a multiple of 180° and (because of the "plus / minus") also 30° less than that same multiple, so (180n)° ± 30° will cover this part. x = (180n)° ± 30°, (180n)° + 45° for all integers n.

  9. 3.3: Solving Trigonometric Equations

    Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. ... Solve the problem exactly: \(2 {\sin}^2 \theta−1=0\), \(0≤\theta<2\pi\).

  10. Trigonometric Equations

    Now on to solving equations. The general method of solving an equation is to convert it into the form of one ratio only. Then, using these results, we can obtain solutions. Solving basic equations can be taken care of with the trigonometric R method. Consider the following example: Solve the following equation: \[\cos x - \sqrt{3}\sin x = 2.\]

  11. Solving Triangles

    1. Angles Add to 180°: A + B + C = 180° When you know two angles you can find the third. 2. Law of Sines (the Sine Rule): a sin (A) = b sin (B) = c sin (C) When there is an angle opposite a side, this equation comes to the rescue. Note: angle A is opposite side a, B is opposite b, and C is opposite c. 3. Law of Cosines (the Cosine Rule):

  12. The 36 Trig Identities You Need to Know

    Knowing key trig identities helps you remember and understand important mathematical principles and solve numerous math problems. The 25 Most Important Trig Identities Below are six categories of trig identities that you'll be seeing often. Each of these is a key trig identity and should be memorized.

  13. How to Solve Trig Identities and Tips on Proving Trigonometric

    STEP 1: Convert all sec, csc, cot, and tan to sin and cos. Most of this can be done using the quotient and reciprocal identities. STEP 2: Check all the angles for sums and differences and use the appropriate identities to remove them. STEP 3: Check for angle multiples and remove them using the appropriate formulas.

  14. How to Solve Trigonometry Problems : 6 Steps

    Step 1: Formulas and Definitions First Slide: Formulas - Sin = Opp/Hyp, Cos = Adj/Hypotenuse Tan = Opp/Adj Note: x is the angle we are using to determine the opp, adj, or the hypotenuse. If it were another angle, then the opposite and adjacent would change. Tricks : Soh Cah Toa. You may have seen or heard of this many times.

  15. Solving Word Problems with Trigonometry

    Correct answer: Explanation: 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. Make sure to round to places after the decimal. The flagpole is feet tall. Report an Error Example Question #1 : Solving Word Problems With Trigonometry

  16. Trigonometric equations and identities

    In this unit, you'll explore the power and beauty of trigonometric equations and identities, which allow you to express and relate different aspects of triangles, circles, and waves. You'll learn how to use trigonometric functions, their inverses, and various identities to solve and check equations and inequalities, and to model and analyze problems involving periodic motion, sound, light, and ...

  17. Trigonometric Equation Calculator

    To solve a trigonometric simplify the equation using trigonometric identities. Then, write the equation in a standard form, and isolate the variable using algebraic manipulation to solve for the variable. Use inverse trigonometric functions to find the solutions, and check for extraneous solutions. What is a basic trigonometric equation?

  18. Trigonometry Problems: Problems with Solutions

    Problem 14 sent by Vasa Shanmukha Reddy. If cot (x) = 2 then find \displaystyle \frac { (2+2\sin x) (1-\sin x)} { (1+\cos x) (2-2\cos x)} (1+cosx)(2 −2cosx)(2+2sinx)(1−sinx) Problem 15. Find the exact value of cos 15°. Problem 16. Calculate sin75°sin15° =. Problem 17. Calculate the exact value of sin15°. Problem 18.

  19. Calculus I

    Most trig equations won't come down to one of those and will in fact need a calculator to solve. The next section is devoted to this kind of problem. In this section we will discuss how to solve trig equations. The answers to the equations in this section will all be one of the "standard" angles that most students have memorized after a trig class.

  20. Trigonometry Word Problems Practice

    1. From a point on the ground 47 feet from the foot of a tree, the angle of elevation of the top of the tree is 35º. Find the height of the tree to the nearest foot. 2. An 8 foot metal guy wire is attached to a broken stop sign to secure its position until repairs can be made.

  21. Calculus I

    Here is a set of practice problems to accompany the Solving Trig Equations Section of the Review chapter of the notes for Paul Dawkins Calculus I course at Lamar University.

  22. Trigonometric functions

    Algebra (all content) Unit 14: Trigonometric functions About this unit 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

  23. Trigonometry Problem Solver

    Problem Solver Subjects. Our math problem solver that lets you input a wide variety of trigonometry math problems and it will provide a step by step answer. This math solver excels at math word problems as well as a wide range of math subjects. Here are example math problems within each subject that can be input into the calculator and solved.

  24. Heavy Machinery Meets AI

    Summary. Until recently most incumbent industrial companies didn't use highly advanced software in their products. But now the sector's leaders have begun applying generative AI and machine ...

  25. Shakti Dhar on Instagram: "#nda #TRIGONOMETRY #upsc #ltgrade #trending

    21 likes, 2 comments - dharshakti on February 22, 2024: "#nda #TRIGONOMETRY #upsc #ltgrade #trending #jeemains #shaktisir @MATHS WITH SHAKTI SIR In thi..." Shakti Dhar on Instagram: "#nda #TRIGONOMETRY #upsc #ltgrade #trending #jeemains #shaktisir @MATHS WITH SHAKTI SIR In this video SHAKTI SIR deals with solving lengthy TRIGONOMETRY problems ...