## Precalculus (6th Edition)

Published by Pearson

# Chapter 7 - Trigonometric Identities and Equations - Chapter 7 Test Prep - Review Exercises - Page 737: 67

#### Answer

$\tan(\frac{x}{2}+\frac{\pi}{4})=\sec x+\tan x$

#### Work Step by Step

Start with the left side: $\tan(\frac{x}{2}+\frac{\pi}{4})$ Rewrite tangent as sine divided by cosine: $=\frac{\sin (\frac{x}{2}+\frac{\pi}{4})}{\cos(\frac{x}{2}+\frac{\pi}{4})}$ Use addition formulas for sine and cosine: $=\frac{\sin\frac{x}{2} \cos\frac{\pi}{4}+\cos\frac{x}{2} \sin\frac{\pi}{4}}{\cos\frac{x}{2} \cos\frac{\pi}{4}-\sin\frac{x}{2} \sin\frac{\pi}{4}}$ Evaluate $\sin\frac{\pi}{4}$ and $\cos\frac{\pi}{4}$: $=\frac{\sin\frac{x}{2}*\frac{\sqrt{2}}{2}+\cos\frac{x}{2}*\frac{\sqrt{2}}{2}}{\cos\frac{x}{2} *\frac{\sqrt{2}}{2}-\sin\frac{x}{2} *\frac{\sqrt{2}}{2}}$ Divide top and bottom by $\frac{\sqrt{2}}{2}$: $=\frac{\sin\frac{x}{2}+\cos\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}$ Multiply top and bottom by $\cos\frac{x}{2}+\sin\frac{x}{2}$: $=\frac{\sin\frac{x}{2}+\cos\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}*\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}+\sin\frac{x}{2}}$ $=\frac{\sin^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}+\cos^2\frac{x}{2}}{\cos^2\frac{x}{2}-\sin\frac{x}{2}}$ $=\frac{(\sin^2\frac{x}{2}+\cos^2\frac{x}{2})+2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac{x}{2}-\sin\frac{x}{2}}$ Use double-angle formulas for sine and cosine to simplify the expression: $=\frac{1+\sin(2*\frac{x}{2})}{\cos (2*\frac{x}{2})}$ $=\frac{1+\sin x}{\cos x}$ $=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$ $=\sec x+\tan x$

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