<meta http-equiv="refresh" content="1; url=/nojavascript/"> Proofs of Trigonometric Identities ( Read ) | Trigonometry | CK-12 Foundation
Skip Navigation
You are viewing an older version of this Concept. Go to the latest version.

Proofs of Trigonometric Identities

Best Score
Practice Proofs of Trigonometric Identities
Best Score
Practice Now

Verifying a Trigonometric Identity

Verify that \frac{\sin^2x}{\tan^2x}=1 - \sin^2x .


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 \frac{\cot^2x}{\csc x}=\csc x - \sin x .

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{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}

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

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, \sin^2 \theta+\cos^2 \theta=1 , and isolated the \cos^2x=1- \sin^2x .

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 \frac{\cos^2x}{\sin^2x} , we could have also substituted the identity \cot^2x=\csc^2x-1 .

Example B

Verify the identity \frac{\sin x}{1- \cos x}=\frac{1+ \cos x}{\sin x} .

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

\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}&=

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

Example C

Verify the identity \sec(-x)=\sec x .

Solution: Change secant to cosine.

\sec(-x)= \frac{1}{\cos \left(-x\right)}

From the Negative Angle Identities, we know that \cos (-x)=\cos x .

&=\frac{1}{\cos x} \\&=\sec x

Concept Problem Revisit Start by simplifying the lefthand side of the equation.

\frac{\sin^2x}{\tan^2x}=\frac{\sin^2x}{\frac{\sin^2x}{\cos^2x}}\\=\cos^2x .

Now simplify the righthand side of the equation. By manipulating the Trigonometric Identity,

\sin^2x + \cos^2x = 1 , we get \cos^2x = 1 - \sin^2x .

\cos^2x =\cos^2x and the equation is verified.

Guided Practice

Verify the following identities.

1. \cos x \sec x=1

2. 2- \sec^2x=1- \tan^2x

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


1. Change secant to cosine.

\cos x \sec x&=\cos \cdot \frac{1}{\cos x} \\&=1

2. Use the identity 1+ \tan^2 \theta=\sec^2 \theta .

2- \sec^2x&=2-(1+ \tan^2x) \\&=2-1- \tan^2x \\&=1- \tan^2x

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

\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


Verify the following identities.

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

Image Attributions


Email Verified
Well done! You've successfully verified the email address .
Please wait...
Please wait...

Original text