## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set - Page 548: 30

#### Answer

$\frac{\sqrt {15}}{8}$

#### Work Step by Step

$sin^{2}t$ + $cos^{2}t$ = 1 sin t = $\frac{7}{8}$ $(\frac{7}{8})^{2}$ + $cos^{2}t$ = 1 $\frac{49}{64}$ + $cos^{2}t$ = 1 $cos^{2}t$ = 1 -$\frac{49}{64}$ $cos^{2}t$ = $\frac{15}{64}$ cos t = $\sqrt \frac{15}{64}$ = $\frac{\sqrt {15}}{8}$

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