Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - 7.1 - Using Fundamental Identities - 7.1 Exercises - Page 514: 57


$cos~θ=\frac{\sqrt 2}{2}$ $sin~θ=±\frac{\sqrt 2}{2}$

Work Step by Step

$sin^2θ+cos^2θ=1$ $cos^2θ=1-sin^2θ$ $\sqrt 2=\sqrt {4-x^2}$ $\sqrt 2=\sqrt {4-4~sin^2θ}$ $\sqrt 2=\sqrt {4(1-sin^2θ)}$ $\sqrt 2=\sqrt {4~cos^2θ}~~$ (Square both sides) $2=4cos^2θ$ $cos^2θ=\frac{1}{2}$ $cos~θ=±\frac{1}{\sqrt 2}=±\frac{\sqrt 2}{2}$ Since $-\frac{\pi}{2}\ltθ\lt\frac{\pi}{2}$, $cos~θ=\frac{\sqrt 2}{2}$ $sin^2θ=1-cos^2θ=1-\frac{1}{2}=\frac{1}{2}$ $sin~θ=±\frac{1}{\sqrt 2}=±\frac{\sqrt 2}{2}$
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