## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Mid-Chapter Check Point - Page 683: 5

#### Answer

See the explanation below.

#### Work Step by Step

The expression on the left side $\frac{1-\cos 2x}{\sin 2x}$ can be further simplified by using the double angle formula, that is $\cos 2\theta =1-{{\sin }^{2}}\theta$ and $\sin 2\theta =2\sin \theta \cos \theta$. Therefore, the expression can be written as: \begin{align} & \frac{1-\cos 2x}{\sin 2x}=\frac{1-\left( 1-2{{\sin }^{2}}x \right)}{2\sin x\cos x} \\ & =\frac{1-1+2{{\sin }^{2}}x}{2\sin x\cos x} \\ & =\frac{2{{\sin }^{2}}x}{2\sin x\cos x} \\ & =\frac{\sin x}{\cos x} \end{align} We know the quotient identity $tanx=\frac{\sin x}{\cos x}$. Applying the identity to the expression, we get: $\frac{\sin x}{\cos x}=\tan x$ Thus, the left side is equal to the right side $\frac{1-\cos 2x}{\sin 2x}=\tan x$.

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