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# 14.2: Using Trigonometric Identities

Difficulty Level: At Grade Created by: CK-12

Objective

Here you’ll learn how to use trigonometric identities to solve and prove trigonometric statements and problems.

Review Queue

Find the exact value of the trig functions below.

1. sinπ3\begin{align*}\sin \frac{\pi}{3}\end{align*}

2. cos(5π6)\begin{align*}\cos \left(- \frac{5 \pi}{6}\right)\end{align*}

3. sin4π\begin{align*}\sin 4 \pi\end{align*}

4. tan7π4\begin{align*}\tan \frac{7 \pi}{4}\end{align*}

5. cos2π3\begin{align*}\cos \frac{2 \pi}{3}\end{align*}

6. tan(2π3)\begin{align*}\tan \left(- \frac{2 \pi}{3}\right)\end{align*}

## Introduction to Trig Identities

Objective

Here you’ll learn about the basic trigonometric identities and how to use them.

Guidance

Trigonometric identities are true for any value of x\begin{align*}x\end{align*} (as long as the value is in the domain). In the Reciprocal Trigonometric Functions concept from the previous chapter, you learned about secant, cosecant, and cotangent, which are all reciprocal functions of sine, cosine and tangent. These functions can be rewritten as the Reciprocal Identities because they are always true.

Reciprocal Identities: cscθ=1sinθsecθ=1cosθcotθ=1tanθ\begin{align*}\csc \theta=\frac{1}{\sin \theta} \quad \sec \theta=\frac{1}{\cos \theta} \quad \cot \theta=\frac{1}{\tan \theta}\end{align*}

Other identities involve the tangent, variations on the Pythagorean Theorem, phase shifts, and negative angles. We will discover them in this concept.

Example A

tanθ=oppositeadjacent\begin{align*}\tan \theta=\frac{opposite}{adjacent}\end{align*}. Show that tanθ=sinθcosθ\begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}\end{align*}.This is the Tangent Identity.

Solution: Whenever we are trying to verify, or prove, an identity, we start with the statement we are trying to prove and work towards the desired answer. In this case, we will start with tanθ=sinθcosθ\begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}\end{align*} and show that it is equivalent to tanθ=oppositeadjacent\begin{align*}\tan \theta=\frac{opposite}{adjacent}\end{align*}. First, rewrite sine and cosine in terms of the ratios of the sides.

Then, rewrite the complex fraction as a division problem and simplify.

We now have what we wanted to prove and we are done. Once you verify an identity, you may use it to verify other identities.

Example B

Show that sin2θ+cos2θ=1\begin{align*}\sin^2 \theta+ \cos^2 \theta=1\end{align*} is a true identity.

Solution: Change the sine and cosine in the equation into the ratios. In this example, we will use y\begin{align*}y\end{align*} as the opposite side, x\begin{align*}x\end{align*} is the adjacent side, and r\begin{align*}r\end{align*} is the hypotenuse (or radius), as in the unit circle.

sin2θ+cos2θ(yr)2+(xr)2y2r2+x2r2y2+x2r2=1=1=1=1

Now, x2+y2=r2\begin{align*}x^2+y^2=r^2\end{align*} from the Pythagorean Theorem. Substitute this in for the numerator of the fraction.

r2r2=1

This is one of the Pythagorean Identities and very useful.

Example C

Verify that sin(π2θ)=cosθ\begin{align*}\sin \left(\frac{\pi}{2}- \theta \right)=\cos \theta\end{align*} by using the graphs of the functions.

Solution: The function y=sin(π2x)\begin{align*}y=\sin \left(\frac{\pi}{2} - x \right)\end{align*} is a phase shift of π2\begin{align*}\frac{\pi}{2}\end{align*} of the sine curve.

The red function above is y=sinx\begin{align*}y=\sin x\end{align*} and the blue is y=cosx\begin{align*}y=\cos x\end{align*}. If we were to shift the sine curve π2\begin{align*}\frac{\pi}{2}\end{align*}, it would overlap perfectly with the cosine curve, thus proving this Cofunction Identity.

Guided Practice

1. Prove the Pythagorean Identity: 1+tan2θ=sec2θ\begin{align*}1+ \tan^2 \theta=\sec^2 \theta\end{align*}

2. Without graphing, show that sin(θ)=sinθ\begin{align*}\sin (- \theta)=- \sin \theta\end{align*}.

1. First, let’s use the Tangent Identity and the Reciprocal Identity to change tangent and secant in terms of sine and cosine.

1+tan2θ1+sin2θcos2θ=sec2θ=1cos2θ

Now, change the 1 into a fraction with a base of cos2θ\begin{align*}\cos^2 \theta\end{align*} and simplify.

1+sin2θcos2θcos2θcos2θ+sin2θcos2θcos2θ+sin2θcos2θ1cos2θ=1cos2θ=1cos2θ=1cos2θ=1cos2θ

In the second to last step, we arrived at the original Pythagorean Identity sin2θ+cos2θ\begin{align*}\sin^2 \theta+\cos^2 \theta\end{align*} in the numerator of the left-hand side. Therefore, we can substitute in 1 for this and the two sides of the equation are the same.

2. First, recall that sinθ=y\begin{align*}\sin \theta=y\end{align*}, where (x,y)\begin{align*}(x, y)\end{align*} is the endpoint of the terminal side of θ\begin{align*}\theta\end{align*} on the unit circle.

Now, if we have sin(θ)\begin{align*}\sin (- \theta)\end{align*}, what is it’s endpoint? Well, the negative sign tells us that the angle is rotated in a clockwise direction, rather than the usual counter-clockwise. By looking at the picture, we see that sin(θ)=y\begin{align*}\sin (- \theta)=-y\end{align*}. Therefore, if sinθ=y\begin{align*}\sin \theta=y\end{align*}, then sinθ=y\begin{align*}- \sin \theta=-y\end{align*} and combining the equations, we have sin(θ)=sinθ\begin{align*}\sin (- \theta)=- \sin \theta\end{align*}.

Vocabulary

Trigonometric Identity
A trigonometric equation that is true for any x\begin{align*}x\end{align*} value (within the domain).
Verify
To prove a trigonometric identity.
Reciprocal Identities
cscθ=1sinθ,secθ=1cosθ,\begin{align*}\csc \theta=\frac{1}{\sin \theta}, \sec \theta=\frac{1}{\cos \theta},\end{align*} and cotθ=1tanθ\begin{align*}\cot \theta=\frac{1}{\tan \theta}\end{align*}
Tangent Identity
tanθ=sinθcosθ\begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}\end{align*}
Cotangent Identity
cotθ=cosθsinθ\begin{align*}\cot \theta=\frac{\cos \theta}{\sin \theta}\end{align*}
Pythagorean Identities
sin2θ+cos2θ=1,1+tan2θ=sec2θ,\begin{align*}\sin^2 \theta+ \cos^2 \theta=1, 1+ \tan^2 \theta=\sec^2 \theta,\end{align*} and 1+cot2θ=csc2θ\begin{align*}1+ \cot^2 \theta=\csc^2 \theta\end{align*}
Cofunction Identities
sin(π2θ)=cosθ,cos(π2θ)=sinθ,\begin{align*}\sin \left(\frac{\pi}{2}- \theta \right)=\cos \theta, \cos \left(\frac{\pi}{2}- \theta \right)=\sin \theta,\end{align*} and tan(π2θ)=cotθ\begin{align*}\tan \left(\frac{\pi}{2} - \theta \right)=\cot \theta\end{align*}
Negative Angle Identities
sin(θ)=sinθ,cos(θ)=cosθ,\begin{align*}\sin (- \theta)=- \sin \theta, \cos(- \theta)=\cos \theta,\end{align*} and tan(θ)=tanθ\begin{align*}\tan (- \theta)=- \tan \theta\end{align*}

Problem Set

1. Show that cotθ=cosθsinθ\begin{align*}\cot \theta=\frac{\cos \theta}{\sin \theta}\end{align*} by following the steps from Example A.
2. Show that tanθ=secθcscθ\begin{align*}\tan \theta=\frac{\sec \theta}{\csc \theta}\end{align*}. Refer to Example A to help you.
3. Show that 1+cot2θ=csc2θ\begin{align*}1+ \cot^2 \theta=\csc^2 \theta\end{align*} by following the steps from #1 in the Guided Practice.
4. Explain why cos(π2θ)=sinθ\begin{align*}\cos \left(\frac{\pi}{2} - \theta \right)=\sin \theta\end{align*} by using the graphs of the two functions.
5. Following the steps from #2 in the Guided Practice, show that cos(θ)=cosθ\begin{align*}\cos (- \theta)=\cos \theta\end{align*}.
6. Explain why tan(θ)=tanθ\begin{align*}\tan (- \theta)=- \tan \theta\end{align*} is true, using the Tangent Identity and the other Negative Angle Identities.

Verify the following identities.

1. cotθsecθ=cscθ\begin{align*}\cot \theta \sec \theta=\csc \theta\end{align*}
2. cosθcotθ=tanθsecθ\begin{align*}\frac{\cos \theta}{\cot \theta}=\frac{\tan \theta}{\sec \theta}\end{align*}
3. sinθcscθ=1\begin{align*}\sin \theta \csc \theta=1\end{align*}
4. cot(θ)=cotθ\begin{align*}\cot(- \theta)=- \cot \theta\end{align*}
5. tanxcscxcosx=1\begin{align*}\tan x \csc x \cos x=1\end{align*}
6. sin2(x)tan2x=cos2x\begin{align*}\frac{\sin^2 \left(-x\right)}{\tan^2 x}= \cos ^2 x\end{align*}

Show that sin2θ+cos2θ=1\begin{align*}\sin^2 \theta+ \cos^2 \theta=1\end{align*} is true for the following values of θ\begin{align*}\theta\end{align*}.

1. π4\begin{align*}\frac{\pi}{4}\end{align*}
2. 2π3\begin{align*}\frac{2 \pi}{3}\end{align*}
3. 7π6\begin{align*}- \frac{7 \pi}{6}\end{align*}
4. Recall that a function is odd if f(x)=f(x)\begin{align*}f(-x)=-f(x)\end{align*} and even if f(x)=f(x)\begin{align*}f(-x)=f(x)\end{align*}. Which of the six trigonometric functions are odd? Which are even?

## Using Trig Identities to Find Exact Trig Values

Objective

Here you'll use the basic trig identities to find exact trig values of angles other than the critical angles.

Guidance

You can use the Pythagorean, Tangent and Reciprocal Identities to find all six trigonometric values for certain angles. Let’s walk through a few examples so that you understand how to do this.

Example A

Given that cosθ=35\begin{align*}\cos \theta=\frac{3}{5}\end{align*} and 0<θ<π2\begin{align*}0 < \theta < \frac{\pi}{2}\end{align*}, find sinθ\begin{align*}\sin \theta\end{align*}.

Solution: Use the Pythagorean Identity to find sinθ\begin{align*}\sin \theta\end{align*}.

sin2θ+cos2θsin2θ+(35)2sin2θsin2θsinθ=1=1=1925=1625=±45

Because θ\begin{align*}\theta\end{align*} is in the first quadrant, we know that sine will be positive. sinθ=45\begin{align*}\sin \theta=\frac{4}{5}\end{align*}

Example B

Find tanθ\begin{align*}\tan \theta\end{align*} of θ\begin{align*}\theta\end{align*} from Example A.

Solution: Use the Tangent Identity to find tanθ\begin{align*}\tan \theta\end{align*}.

tanθ=sinθcosθ=4535=43

Example C

Find the other three trigonometric functions of θ\begin{align*}\theta\end{align*} from Example.

Solution: To find secant, cosecant, and cotangent use the Reciprocal Identities.

cscθ=1sinθ=145=54secθ=1cosθ=135=53cotθ=1tanθ=143=34

Guided Practice

Find the values of the other five trigonometric functions.

1. tanθ=512,π2<θ<π\begin{align*}\tan \theta=- \frac{5}{12}, \frac{\pi}{2} < \theta < \pi\end{align*}

2. cscθ=8,π<θ<3π2\begin{align*}\csc \theta=-8, \pi < \theta < \frac{3 \pi}{2}\end{align*}

1. First, we know that θ\begin{align*}\theta\end{align*} is in the second quadrant, making sine positive and cosine negative. For this problem, we will use the Pythagorean Identity 1+tan2θ=sec2θ\begin{align*}1+ \tan^2 \theta=\sec^2 \theta\end{align*} to find secant.

1+(512)21+25144169144±13121312=sec2θ=sec2θ=sec2θ=secθ=secθ

If secθ=1312\begin{align*}\sec \theta=- \frac{13}{12}\end{align*}, then cosθ=1213\begin{align*}\cos \theta= - \frac{12}{13}\end{align*}. sinθ=513\begin{align*}\sin \theta=\frac{5}{13}\end{align*} because the numerator value of tangent is the sine and it has the same denominator value as cosine. cscθ=135\begin{align*}\csc \theta=\frac{13}{5}\end{align*} and cotθ=125\begin{align*}\cot \theta=- \frac{12}{5}\end{align*} from the Reciprocal Identities.

2. θ\begin{align*}\theta\end{align*} is in the third quadrant, so both sine and cosine are negative. The reciprocal of cscθ=8\begin{align*}\csc \theta=-8\end{align*}, will give us sinθ=18\begin{align*}\sin \theta=- \frac{1}{8}\end{align*}. Now, use the Pythagorean Identity sin2θ+cos2θ=1\begin{align*}\sin^2 \theta + \cos^2 \theta=1\end{align*} to find cosine.

(18)2+cos2θcos2θcos2θcosθcosθ=1=1164=6364=±378=378

secθ=837=8721,tanθ=137=721,\begin{align*}\sec \theta=- \frac{8}{3 \sqrt{7}}=- \frac{8 \sqrt{7}}{21}, \tan \theta= \frac{1}{3 \sqrt{7}}= \frac{\sqrt{7}}{21},\end{align*} and cotθ=37\begin{align*}\cot \theta=3 \sqrt{7}\end{align*}

Problem Set

1. In which quadrants is the sine value positive? Negative?
2. In which quadrants is the cosine value positive? Negative?
3. In which quadrants is the tangent value positive? Negative?

Find the values of the other five trigonometric functions of θ\begin{align*}\theta\end{align*}.

1. sinθ=817,0<θ<π2\begin{align*}\sin \theta=\frac{8}{17},0 < \theta < \frac{\pi}{2}\end{align*}
2. cosθ=56,π2<θ<π\begin{align*}\cos \theta=- \frac{5}{6}, \frac{\pi}{2} < \theta < \pi\end{align*}
3. tanθ=34,0<θ<π2\begin{align*}\tan \theta= \frac{\sqrt{3}}{4},0 < \theta < \frac{\pi}{2}\end{align*}
4. secθ=419,π<θ<3π2\begin{align*}\sec \theta=- \frac{41}{9}, \pi < \theta < \frac{3 \pi}{2}\end{align*}
5. sinθ=1114,3π2<θ<2π\begin{align*}\sin \theta=- \frac{11}{14}, \frac{3 \pi}{2} < \theta < 2 \pi\end{align*}
6. cosθ=224,0<θ<π2\begin{align*}\cos \theta=\frac{2 \sqrt{2}}{4},0 < \theta < \frac{\pi}{2}\end{align*}
7. cotθ=5,π<θ<3π2\begin{align*}\cot \theta=- \sqrt{5}, \pi < \theta < \frac{3 \pi}{2}\end{align*}
8. cscθ=4,π2<θ<π\begin{align*}\csc \theta=4, \frac{\pi}{2} < \theta < \pi\end{align*}
9. tanθ=710,3π2<θ<2π\begin{align*}\tan \theta=- \frac{7}{10}, \frac{3 \pi}{2} < \theta < 2 \pi\end{align*}
10. Aside from using the identities, how else can you find the values of the other five trigonometric functions?
11. Given that cosθ=611\begin{align*}\cos \theta=\frac{6}{11}\end{align*} and θ\begin{align*}\theta\end{align*} is in the 2nd\begin{align*}2^{nd}\end{align*} quadrant, what is sin(θ)\begin{align*}\sin(- \theta)\end{align*}?
12. Given that tanθ=58\begin{align*}\tan \theta=- \frac{5}{8}\end{align*} and θ\begin{align*}\theta\end{align*} is in the 4th\begin{align*}4^{th}\end{align*} quadrant, what is sec(θ)\begin{align*}\sec(- \theta)\end{align*}?

## Simplifying Trigonometric Expressions

Objective

Here you'll use the basic trig identities to simplify more complicated expressions.

Guidance

Now that you are more familiar with trig identities, we can use them to simplify expressions. Remember, that you can use any of the identities in the Introduction to Trig Identities concept. Here is a list of the identities again:

Reciprocal Identities: cscθ=1sinθ,secθ=1cosθ,\begin{align*}\csc \theta=\frac{1}{\sin \theta}, \sec \theta=\frac{1}{\cos \theta},\end{align*} and cotθ=1tanθ\begin{align*}\cot \theta=\frac{1}{\tan \theta}\end{align*}

Tangent and Cotangent Identities: tanθ=sinθcosθ\begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}\end{align*} and cotθ=cosθsinθ\begin{align*}\cot \theta=\frac{\cos \theta}{\sin \theta}\end{align*}

Pythagorean Identities: sin2θ+cos2θ=1,1+tan2θ=sec2θ,\begin{align*}\sin^2 \theta+ \cos^2 \theta=1, 1+ \tan^2 \theta=\sec^2 \theta,\end{align*} and 1+cot2θ=csc2θ\begin{align*}1+ \cot^2 \theta=\csc^2 \theta\end{align*}

Cofunction Identities: sin(π2θ)=cosθ,cos(π2θ)=sinθ,\begin{align*}\sin \left(\frac{\pi}{2} - \theta\right)=\cos \theta, \cos \left(\frac{\pi}{2} - \theta\right)=\sin \theta,\end{align*} and tan(π2θ)=cotθ\begin{align*}\tan \left(\frac{\pi}{2} - \theta\right)=\cot \theta\end{align*}

Negative Angle Identities: sin(θ)=sinθ,cos(θ)=cosθ,\begin{align*}\sin(- \theta)=- \sin \theta, \cos(- \theta)=\cos \theta,\end{align*} and tan(θ)=tanθ\begin{align*}\tan(- \theta)=- \tan \theta\end{align*}

Example A

Simplify secxsecx1\begin{align*}\frac{\sec x}{\sec x-1}\end{align*}.

Solution: When simplifying trigonometric expressions, one approach is to change everything into sine or cosine. First, we can change secant to cosine using the Reciprocal Identity.

secxsecx11cosx1cosx1

Now, combine the denominator into one fraction by multiplying 1 by cosxcosx\begin{align*}\frac{\cos x}{\cos x}\end{align*}.

1cosx1cosx11cosx1cosxcosxcosx1cosx1cosxcosx

Change this problem into a division problem and simplify.

1cosx1cosxcosx1cosx÷1cosxcosx1cosxcosx1cosx11cosx

Example B

Simplify sin4xcos4xsin2xcos2x\begin{align*}\frac{\sin^4x- \cos^4x}{\sin^2x- \cos^2x}\end{align*}.

Solution: With this problem, we need to factor the numerator and denominator and see if anything cancels. The rules of factoring a quadratic and the special quadratic formulas can be used in this scenario.

In the last step, we simplified to the left hand side of the Pythagorean Identity. Therefore, this expression simplifies to 1.

Example C

Simplify \begin{align*}\sec \theta \tan^2 \theta+\sec \theta\end{align*}.

Solution: First, pull out the GCF.

\begin{align*}\sec \theta \tan^2 \theta+ \sec \theta \rightarrow \sec \theta(\tan^2 \theta+1)\end{align*}

Now, \begin{align*}\tan^2 \theta+1=\sec^2 \theta\end{align*} from the Pythagorean Identities, so simplify further.

\begin{align*}\sec \theta(\tan^2 \theta+1) \rightarrow \sec \theta \cdot \sec^2 \theta \rightarrow \sec^3 \theta\end{align*}

Guided Practice

Simplify the following trigonometric expressions.

1. \begin{align*}\cos \left(\frac{\pi}{2} - x\right) \cot x\end{align*}

2. \begin{align*}\frac{\sin \left(-x\right) \cos x}{\tan x}\end{align*}

3. \begin{align*}\frac{\cot x \cos x}{\tan \left(-x\right) \sin \left(\frac{\pi}{2}-x \right)}\end{align*}

1. Use the Cotangent Identity and the Cofunction Identity \begin{align*}\cos \left(\frac{\pi}{2}- \theta \right)=\sin \theta\end{align*}.

\begin{align*}\cos \left(\frac{\pi}{2}-x\right) \cot x \rightarrow \cancel{\sin x} \cdot \frac{\cos x}{\cancel{\sin x}} \rightarrow \cos x\end{align*}

2. Use the Negative Angle Identity and the Tangent Identity.

\begin{align*}\frac{\sin \left(-x\right) \cos x}{\tan x} \rightarrow \frac{- \sin x \cos x}{\frac{\sin x}{\cos x}} \rightarrow - \cancel{\sin x} \cos x \cdot \frac{\cos x}{\sin x} \rightarrow - \cos^2x\end{align*}

3. In this problem, you will use several identities.

\begin{align*}\frac{\cot x \cos x}{\tan \left(-x\right) \sin \left(\frac{\pi}{2}-x\right)} \rightarrow \frac{\frac{\cos x}{\sin x} \cdot \cos x}{- \frac{\sin x}{\cancel{\cos x}} \cdot \cancel{\cos x}} \rightarrow \frac{\frac{\cos^2 x}{\sin x}}{- \sin x} \rightarrow \frac{\cos^2x}{\sin x} \cdot - \frac{1}{\sin x} \rightarrow - \frac{\cos^2x}{\sin^2 x} \rightarrow - \cot^2x\end{align*}

Problem Set

Simplify the following expressions.

1. \begin{align*}\cot x \sin x\end{align*}
2. \begin{align*}\cos^2x \tan(-x)\end{align*}
3. \begin{align*}\frac{\cos \left(-x\right)}{\sin \left(-x\right)}\end{align*}
4. \begin{align*}\sec x \cos(-x)- \sin^2x\end{align*}
5. \begin{align*}\sin x(1+ \cot^2x)\end{align*}
6. \begin{align*}1- \sin^2 \left(\frac{\pi}{2} - x\right)\end{align*}
7. \begin{align*}1- \cos^2 \left(\frac{\pi}{2}-x\right)\end{align*}
8. \begin{align*}\frac{\tan \left(\frac{\pi}{2}-x\right) \sec x}{1- \csc^2 x}\end{align*}
9. \begin{align*}\frac{\cos^2x \tan^2 \left(-x\right)-1}{\cos^2x}\end{align*}
10. \begin{align*}\cot^2x+ \sin^2x+ \cos^2(-x)\end{align*}
11. \begin{align*}\frac{\sec x \sin x+ \cos \left(\frac{\pi}{2}-x\right)}{1+ \sin \left(-x\right)}\end{align*}
12. \begin{align*}\frac{\cos \left(-x\right)}{1+ \sin \left(-x\right)}\end{align*}
13. \begin{align*}\frac{\sin^2 \left(-x\right)}{\tan^2x}\end{align*}
14. \begin{align*}\tan \left(\frac{\pi}{2}-x\right) \cot x- \csc^2 x\end{align*}
15. \begin{align*}\frac{\csc x \left(1- \cos^2x \right)}{\sin x \cos x}\end{align*}

## Verifying a Trigonometry Identity

Objective

Here you'll use the basic trig identities to prove other identities.

Guidance

This concept continues where the previous one left off. Now that you are comfortable simplifying expressions, we will extend the idea to verifying entire identities. Here are a few helpful hints to verify an identity:

• Change everything into terms of sine and cosine.
• Use the identities when you can.
• Start with simplifying the left-hand side of the equation, then, once you get stuck, simplify the right-hand side. As long as the two sides end up with the same final expression, the identity is true.

Example A

Verify the identity \begin{align*}\frac{\cot^2x}{\csc x}=\csc x - \sin x\end{align*}.

Solution: Rather than have an equal sign between the two sides of the equation, we will draw a vertical line so that it is easier to see what we do to each side of the equation. Start with changing everything into sine and cosine.

\begin{align*}\begin{array}{c|c c} \frac{\cot^2x}{\csc x} & \csc x - \sin x \\ \frac{\frac{\cos^2x}{\sin^2x}}{\frac{1}{\sin x}} & \frac{1}{\sin x}- \sin x \\ \frac{\cos^2x}{\sin x}\end{array}\end{align*}

Now, it looks like we are at an impasse with the left-hand side. Let’s combine the right-hand side by giving them same denominator.

\begin{align*}\begin{array}{|c} \frac{1}{\sin x}- \frac{\sin^2x}{\sin x} \\ \frac{1- \sin^2x}{\sin x} \\ \frac{\cos^2x}{\sin x}\end{array}\end{align*}

The two sides reduce to the same expression, so we can conclude this is a valid identity. In the last step, we used the Pythagorean Identity, \begin{align*}\sin^2 \theta+\cos^2 \theta=1\end{align*}, and isolated the \begin{align*}\cos^2x=1- \sin^2x\end{align*}.

There are usually more than one way to verify a trig identity. When proving this identity in the first step, rather than changing the cotangent to \begin{align*}\frac{\cos^2x}{\sin^2x}\end{align*}, we could have also substituted the identity \begin{align*}\cot^2x=\csc^2x-1\end{align*}.

Example B

Verify the identity \begin{align*}\frac{\sin x}{1- \cos x}=\frac{1+ \cos x}{\sin x}\end{align*}.

Solution: Multiply the left-hand side of the equation by \begin{align*}\frac{1+ \cos x}{1+ \cos x}\end{align*}.

\begin{align*}\frac{\sin x}{1- \cos x}&= \frac{1+ \cos x}{\sin x} \\ \frac{1+ \cos x}{1+ \cos x} \cdot \frac{\sin x}{1- \cos x}&= \\ \frac{\sin \left(1+\cos x\right)}{1- \cos^2x}&= \\ \frac{\sin \left(1+\cos x\right)}{\sin^2x}&= \\ \frac{1+\cos x}{\sin x}&=\end{align*}

The two sides are the same, so we are done.

Example C

Verify the identity \begin{align*}\sec(-x)=\sec x\end{align*}.

Solution: Change secant to cosine.

\begin{align*}\sec(-x)= \frac{1}{\cos \left(-x\right)}\end{align*}

From the Negative Angle Identities, we know that \begin{align*}\cos (-x)=\cos x\end{align*}.

\begin{align*}&=\frac{1}{\cos x} \\ &=\sec x\end{align*}

Guided Practice

Verify the following identities.

1. \begin{align*}\cos x \sec x=1\end{align*}

2. \begin{align*}2- \sec^2x=1- \tan^2x\end{align*}

3. \begin{align*}\frac{\cos \left(-x\right)}{1+ \sin \left(-x\right)}=\sec x+ \tan x\end{align*}

1. Change secant to cosine.

\begin{align*}\cos x \sec x&=\cos \cdot \frac{1}{\cos x} \\ &=1\end{align*}

2. Use the identity \begin{align*}1+ \tan^2 \theta=\sec^2 \theta\end{align*}.

\begin{align*}2- \sec^2x&=2-(1+ \tan^2x) \\ &=2-1- \tan^2x \\ &=1- \tan^2x\end{align*}

3. Here, start with the Negative Angle Identities and multiply the top and bottom by \begin{align*}\frac{1+ \sin x}{1+ \sin x}\end{align*} to make the denominator a monomial.

\begin{align*}\frac{\cos \left(-x\right)}{1+ \sin \left(-x\right)}&=\frac{\cos x}{1- \sin x} \cdot \frac{1+ \sin x}{1+ \sin x} \\ &=\frac{\cos x \left(1+ \sin x \right)}{1- \sin^2x} \\ &=\frac{\cos x \left(1+ \sin x\right)}{\cos^2x} \\ &=\frac{1+ \sin x}{\cos x} \\ &=\frac{1}{\cos x}+ \frac{\sin x}{\cos x} \\ &=\sec x+ \tan x\end{align*}

Problem Set

Verify the following identities.

1. \begin{align*}\cot (-x)=- \cot x\end{align*}
2. \begin{align*}\csc (-x)=- \csc x\end{align*}
3. \begin{align*}\tan x \csc x \cos x=1\end{align*}
4. \begin{align*}\sin x+ \cos x \cot x=\csc x\end{align*}
5. \begin{align*}\csc \left(\frac{\pi}{2}-x\right)=\sec x\end{align*}
6. \begin{align*}\tan \left(\frac{\pi}{2}-x\right)=\tan x\end{align*}
7. \begin{align*}\frac{\csc x}{\sin x}- \frac{\cot x}{\tan x}=1\end{align*}
8. \begin{align*}\frac{\tan^2x}{\tan^2x+1}=\sin^2x\end{align*}
9. \begin{align*}(\sin x- \cos x)^2+(\sin x+ \cos x)^2=2\end{align*}
10. \begin{align*}\sin x- \sin x \cos^2x= \sin^3x\end{align*}
11. \begin{align*}\tan^2x+1+\tan x \sec x=\frac{1+ \sin x}{\cos^2x}\end{align*}
12. \begin{align*}\cos^2x=\frac{\csc x \cos x}{\tan x+ \cot x}\end{align*}
13. \begin{align*}\frac{1}{1- \sin x} - \frac{1}{1+ \sin x}=2 \tan x \sec x\end{align*}
14. \begin{align*}\csc^4x- \cot^4x=\csc^2x+\cot^2x\end{align*}
15. \begin{align*}(\sin x - \tan x)(\cos x- \cot x)=(\sin x-1)(\cos x-1)\end{align*}

## Date Created:

Apr 23, 2013

Jun 11, 2015
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