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# Proofs of Trigonometric Identities

## Convert to sine/cosine, use basic identities, and simplify sides of the equation.

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Practice Proofs of Trigonometric Identities
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Verifying a Trigonometric Identity

Verify that sin2xtan2x=1sin2x$\frac{\sin^2x}{\tan^2x}=1 - \sin^2x$ .

### 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 cot2xcscx=cscxsinx$\frac{\cot^2x}{\csc x}=\csc x - \sin x$ .

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.

cot2xcscxcos2xsin2x1sinxcos2xsinxcscxsinx1sinxsinx$\begin{array}{c|c c} \frac{\cot^2x}{\csc x} & \csc x - \sin x \\ \frac{\frac{\cos^2x}{\sin^2x}}{\frac{1}{\sin x}} & \frac{1}{\sin x}- \sin x \\ \frac{\cos^2x}{\sin x}\end{array}$

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.

1sinxsin2xsinx1sin2xsinxcos2xsinx$\begin{array}{|c} \frac{1}{\sin x}- \frac{\sin^2x}{\sin x} \\ \frac{1- \sin^2x}{\sin x} \\ \frac{\cos^2x}{\sin x}\end{array}$

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, sin2θ+cos2θ=1$\sin^2 \theta+\cos^2 \theta=1$ , and isolated the cos2x=1sin2x$\cos^2x=1- \sin^2x$ .

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 cos2xsin2x$\frac{\cos^2x}{\sin^2x}$ , we could have also substituted the identity cot2x=csc2x1$\cot^2x=\csc^2x-1$ .

#### Example B

Verify the identity sinx1cosx=1+cosxsinx$\frac{\sin x}{1- \cos x}=\frac{1+ \cos x}{\sin x}$ .

Solution: Multiply the left-hand side of the equation by 1+cosx1+cosx$\frac{1+ \cos x}{1+ \cos x}$ .

\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)=secx$\sec(-x)=\sec x$ .

Solution: Change secant to cosine.

sec(x)=1cos(x)

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

&=\frac{1}{\cos x} \\
&=\sec x

Concept Problem Revisit Start by simplifying the left-hand side of the equation.

sin2xtan2x=sin2xsin2xcos2x=cos2x
.

Now simplify the right-hand side of the equation. By manipulating the Trigonometric Identity,

sin2x+cos2x=1$\sin^2x + \cos^2x = 1$ , we get cos2x=1sin2x$\cos^2x = 1 - \sin^2x$ .

cos2x=cos2x$\cos^2x =\cos^2x$ and the equation is verified.

### Guided Practice

Verify the following identities.

1. cosxsecx=1$\cos x \sec x=1$

2. 2sec2x=1tan2x$2- \sec^2x=1- \tan^2x$

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

1. Change secant to cosine.

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

2. Use the identity 1+tan2θ=sec2θ$1+ \tan^2 \theta=\sec^2 \theta$ .

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 1+sinx1+sinx$\frac{1+ \sin x}{1+ \sin x}$ 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

### Explore More

Verify the following identities.

1. cot(x)=cotx$\cot (-x)=- \cot x$
2. csc(x)=cscx$\csc (-x)=- \csc x$
3. tanxcscxcosx=1$\tan x \csc x \cos x=1$
4. sinx+cosxcotx=cscx$\sin x+ \cos x \cot x=\csc x$
5. csc(π2x)=secx$\csc \left(\frac{\pi}{2}-x\right)=\sec x$
6. tan(π2x)=tanx$\tan \left(\frac{\pi}{2}-x\right)=\tan x$
7. cscxsinxcotxtanx=1$\frac{\csc x}{\sin x}- \frac{\cot x}{\tan x}=1$
8. tan2xtan2x+1=sin2x$\frac{\tan^2x}{\tan^2x+1}=\sin^2x$
9. (sinxcosx)2+(sinx+cosx)2=2$(\sin x- \cos x)^2+(\sin x+ \cos x)^2=2$
10. sinxsinxcos2x=sin3x$\sin x- \sin x \cos^2x= \sin^3x$
11. tan2x+1+tanxsecx=1+sinxcos2x$\tan^2x+1+\tan x \sec x=\frac{1+ \sin x}{\cos^2x}$
12. cos2x=cscxcosxtanx+cotx$\cos^2x=\frac{\csc x \cos x}{\tan x+ \cot x}$
13. 11sinx11+sinx=2tanxsecx$\frac{1}{1- \sin x} - \frac{1}{1+ \sin x}=2 \tan x \sec x$
14. csc4xcot4x=csc2x+cot2x$\csc^4x- \cot^4x=\csc^2x+\cot^2x$
15. (sinxtanx)(cosxcotx)=(sinx1)(cosx1)$(\sin x - \tan x)(\cos x- \cot x)=(\sin x-1)(\cos x-1)$