<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation

14.2: Using Trigonometric Identities

Difficulty Level: At Grade Created by: CK-12
Turn In


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

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

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

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

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

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

Introduction to Trig Identities


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


Trigonometric identities are true for any value of \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: \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

\begin{align*}\tan \theta=\frac{opposite}{adjacent}\end{align*}. Show that \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 \begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}\end{align*} and show that it is equivalent to \begin{align*}\tan \theta=\frac{opposite}{adjacent}\end{align*}. First, rewrite sine and cosine in terms of the ratios of the sides.

\begin{align*}\tan \theta&=\frac{\sin \theta}{\cos \theta} \\ &=\frac{\frac{opposite}{hypotenuse}}{\frac{adjacent}{hypotenuse}}\end{align*}

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

\begin{align*}&=\frac{opposite}{hypotenuse} \div \frac{adjacent}{hypotenuse} \\ &=\frac{opposite}{\cancel{hypotenuse}} \cdot \frac{\cancel{hypotenuse}}{adjacent} \\ &=\frac{opposite}{adjacent}\end{align*}

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 \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 \begin{align*}y\end{align*} as the opposite side, \begin{align*}x\end{align*} is the adjacent side, and \begin{align*}r\end{align*} is the hypotenuse (or radius), as in the unit circle.

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

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


This is one of the Pythagorean Identities and very useful.

Example C

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

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

The red function above is \begin{align*}y=\sin x\end{align*} and the blue is \begin{align*}y=\cos x\end{align*}. If we were to shift the sine curve \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: \begin{align*}1+ \tan^2 \theta=\sec^2 \theta\end{align*}

2. Without graphing, show that \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.

\begin{align*}1+ \tan^2 \theta&=\sec^2 \theta \\ 1+ \frac{\sin^2 \theta}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta}\end{align*}

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

\begin{align*}1+ \frac{\sin^2 \theta}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta} \\ \frac{\cos^2 \theta}{\cos^2 \theta}+ \frac{\sin^2 \theta}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta} \\ \frac{\cos^2 \theta+ \sin^2 \theta}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta} \\ \frac{1}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta}\end{align*}

In the second to last step, we arrived at the original Pythagorean Identity \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 \begin{align*}\sin \theta=y\end{align*}, where \begin{align*}(x, y)\end{align*} is the endpoint of the terminal side of \begin{align*}\theta\end{align*} on the unit circle.

Now, if we have \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 \begin{align*}\sin (- \theta)=-y\end{align*}. Therefore, if \begin{align*}\sin \theta=y\end{align*}, then \begin{align*}- \sin \theta=-y\end{align*} and combining the equations, we have \begin{align*}\sin (- \theta)=- \sin \theta\end{align*}.


Trigonometric Identity
A trigonometric equation that is true for any \begin{align*}x\end{align*} value (within the domain).
To prove a trigonometric identity.
Reciprocal Identities
\begin{align*}\csc \theta=\frac{1}{\sin \theta}, \sec \theta=\frac{1}{\cos \theta},\end{align*} and \begin{align*}\cot \theta=\frac{1}{\tan \theta}\end{align*}
Tangent Identity
\begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}\end{align*}
Cotangent Identity
\begin{align*}\cot \theta=\frac{\cos \theta}{\sin \theta}\end{align*}
Pythagorean Identities
\begin{align*}\sin^2 \theta+ \cos^2 \theta=1, 1+ \tan^2 \theta=\sec^2 \theta,\end{align*} and \begin{align*}1+ \cot^2 \theta=\csc^2 \theta\end{align*}
Cofunction Identities
\begin{align*}\sin \left(\frac{\pi}{2}- \theta \right)=\cos \theta, \cos \left(\frac{\pi}{2}- \theta \right)=\sin \theta,\end{align*} and \begin{align*}\tan \left(\frac{\pi}{2} - \theta \right)=\cot \theta\end{align*}
Negative Angle Identities
\begin{align*}\sin (- \theta)=- \sin \theta, \cos(- \theta)=\cos \theta,\end{align*} and \begin{align*}\tan (- \theta)=- \tan \theta\end{align*}

Problem Set

  1. Show that \begin{align*}\cot \theta=\frac{\cos \theta}{\sin \theta}\end{align*} by following the steps from Example A.
  2. Show that \begin{align*}\tan \theta=\frac{\sec \theta}{\csc \theta}\end{align*}. Refer to Example A to help you.
  3. Show that \begin{align*}1+ \cot^2 \theta=\csc^2 \theta\end{align*} by following the steps from #1 in the Guided Practice.
  4. Explain why \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 \begin{align*}\cos (- \theta)=\cos \theta\end{align*}.
  6. Explain why \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. \begin{align*}\cot \theta \sec \theta=\csc \theta\end{align*}
  2. \begin{align*}\frac{\cos \theta}{\cot \theta}=\frac{\tan \theta}{\sec \theta}\end{align*}
  3. \begin{align*}\sin \theta \csc \theta=1\end{align*}
  4. \begin{align*}\cot(- \theta)=- \cot \theta\end{align*}
  5. \begin{align*}\tan x \csc x \cos x=1\end{align*}
  6. \begin{align*}\frac{\sin^2 \left(-x\right)}{\tan^2 x}= \cos ^2 x\end{align*}

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

  1. \begin{align*}\frac{\pi}{4}\end{align*}
  2. \begin{align*}\frac{2 \pi}{3}\end{align*}
  3. \begin{align*}- \frac{7 \pi}{6}\end{align*}
  4. Recall that a function is odd if \begin{align*}f(-x)=-f(x)\end{align*} and even if \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


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


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 \begin{align*}\cos \theta=\frac{3}{5}\end{align*} and \begin{align*}0 < \theta < \frac{\pi}{2}\end{align*}, find \begin{align*}\sin \theta\end{align*}.

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

\begin{align*}\sin^2 \theta+\cos^2 \theta&=1 \\ \sin^2 \theta+ \left(\frac{3}{5}\right)^2&=1 \\ \sin^2 \theta&=1- \frac{9}{25} \\ \sin^2 \theta&=\frac{16}{25} \\ \sin \theta&= \pm \frac{4}{5}\end{align*}

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

Example B

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

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

\begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\end{align*}

Example C

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

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

\begin{align*}\csc \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{4}{5}}= \frac{5}{4} \quad \sec \theta=\frac{1}{\cos \theta}=\frac{1}{\frac{3}{5}}=\frac{5}{3} \quad \cot \theta =\frac{1}{\tan \theta}=\frac{1}{\frac{4}{3}}=\frac{3}{4}\end{align*}

Guided Practice

Find the values of the other five trigonometric functions.

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

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


1. First, we know that \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 \begin{align*}1+ \tan^2 \theta=\sec^2 \theta\end{align*} to find secant.

\begin{align*}1+ \left(- \frac{5}{12}\right)^2&=\sec^2 \theta \\ 1+ \frac{25}{144}&=\sec^2 \theta \\ \frac{169}{144}&=\sec^2 \theta \\ \pm \frac{13}{12}&=\sec \theta \\ - \frac{13}{12}&=\sec \theta\end{align*}

If \begin{align*}\sec \theta=- \frac{13}{12}\end{align*}, then \begin{align*}\cos \theta= - \frac{12}{13}\end{align*}. \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. \begin{align*}\csc \theta=\frac{13}{5}\end{align*} and \begin{align*}\cot \theta=- \frac{12}{5}\end{align*} from the Reciprocal Identities.

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

\begin{align*}\left(- \frac{1}{8}\right)^2+ \cos^2 \theta&=1 \\ \cos^2 \theta&=1- \frac{1}{64} \\ \cos^2 \theta&=\frac{63}{64} \\ \cos \theta&=\pm \frac{3 \sqrt{7}}{8} \\ \cos \theta&=- \frac{3 \sqrt{7}}{8} \\ \end{align*}

\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 \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 \begin{align*}\theta\end{align*}.

  1. \begin{align*}\sin \theta=\frac{8}{17},0 < \theta < \frac{\pi}{2}\end{align*}
  2. \begin{align*}\cos \theta=- \frac{5}{6}, \frac{\pi}{2} < \theta < \pi\end{align*}
  3. \begin{align*}\tan \theta= \frac{\sqrt{3}}{4},0 < \theta < \frac{\pi}{2}\end{align*}
  4. \begin{align*}\sec \theta=- \frac{41}{9}, \pi < \theta < \frac{3 \pi}{2}\end{align*}
  5. \begin{align*}\sin \theta=- \frac{11}{14}, \frac{3 \pi}{2} < \theta < 2 \pi\end{align*}
  6. \begin{align*}\cos \theta=\frac{2 \sqrt{2}}{4},0 < \theta < \frac{\pi}{2}\end{align*}
  7. \begin{align*}\cot \theta=- \sqrt{5}, \pi < \theta < \frac{3 \pi}{2}\end{align*}
  8. \begin{align*}\csc \theta=4, \frac{\pi}{2} < \theta < \pi\end{align*}
  9. \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 \begin{align*}\cos \theta=\frac{6}{11}\end{align*} and \begin{align*}\theta\end{align*} is in the \begin{align*}2^{nd}\end{align*} quadrant, what is \begin{align*}\sin(- \theta)\end{align*}?
  12. Given that \begin{align*}\tan \theta=- \frac{5}{8}\end{align*} and \begin{align*}\theta\end{align*} is in the \begin{align*}4^{th}\end{align*} quadrant, what is \begin{align*}\sec(- \theta)\end{align*}?

Simplifying Trigonometric Expressions


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


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: \begin{align*}\csc \theta=\frac{1}{\sin \theta}, \sec \theta=\frac{1}{\cos \theta},\end{align*} and \begin{align*}\cot \theta=\frac{1}{\tan \theta}\end{align*}

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

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

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

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

Example A

Simplify \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.

\begin{align*}\frac{\sec x}{\sec x - 1} \rightarrow \frac{\frac{1}{\cos x}}{\frac{1}{\cos x}-1}\end{align*}

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

\begin{align*}\frac{\frac{1}{\cos x}}{\frac{1}{\cos x}-1} \rightarrow \frac{\frac{1}{\cos x}}{\frac{1}{\cos x}- \frac{\cos x}{\cos x}} \rightarrow \frac{\frac{1}{\cos x}}{\frac{1- \cos x}{\cos x}}\end{align*}

Change this problem into a division problem and simplify.

\begin{align*}\frac{\frac{1}{\cos x}}{\frac{1-\cos x}{\cos x}} \rightarrow & \frac{1}{\cos x} \div \frac{1- \cos x}{\cos x} \\ & \frac{1}{\cancel{\cos x}} \cdot \frac{\cancel{\cos x}}{1- \cos x} \\ & \frac{1}{1- \cos x}\end{align*}

Example B

Simplify \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.

\begin{align*}\frac{\sin^4x - \cos^4x}{\sin^2x-\cos^2x} \rightarrow \frac{\cancel{\left(\sin^2x-\cos^2x\right)} \left(\sin^2x+ \cos^2x\right)}{\cancel{\left(\sin^2x-\cos^2x\right)}} \rightarrow \sin^2x+\cos^2x \rightarrow 1\end{align*}

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


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


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*}

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

Image Attributions

Show Hide Details
Date Created:
Apr 23, 2013
Last Modified:
Jul 06, 2016
Save or share your relevant files like activites, homework and worksheet.
To add resources, you must be the owner of the section. Click Customize to make your own copy.
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original