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

### Verifying Trigonometric Identities

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.

Let's verify the following identities.

1. cot2xcscx=cscxsinx\begin{align*}\frac{\cot^2x}{\csc x}=\csc x - \sin x\end{align*}

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{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*}.

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

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.

1. \begin{align*}\sec(-x)=\sec x\end{align*}

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*}

### Examples

#### Example 1

Earlier, you were asked to verify that \begin{align*}\frac{\sin^2x}{\tan^2x}=1 - \sin^2x\end{align*}

Start by simplifying the left-hand side of the equation.

\begin{align*}\frac{\sin^2x}{\tan^2x}=\frac{\sin^2x}{\frac{\sin^2x}{\cos^2x}}\\ =\cos^2x\end{align*}

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

\begin{align*}\sin^2x + \cos^2x = 1\end{align*}, we get \begin{align*}\cos^2x = 1 - \sin^2x\end{align*}.

\begin{align*}\cos^2x =\cos^2x\end{align*} and the equation is verified.

Verify the following identities.

#### Example 2

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

Change secant to cosine.

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

#### Example 3

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

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*}

#### Example 4

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

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*}

### Review

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*}

To see the Review answers, open this PDF file and look for section 14.9.

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