Algebra and Trigonometry 10th Edition

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

Chapter 6 - Review Questions - Page 501: 55


$sin~495°=\frac{\sqrt 2}{2}$ $cos~495°=-\frac{\sqrt 2}{2}$ $tan~495°=-1$

Work Step by Step

Let $θ'$ be the reference angle. We must have that $0°\leqθ'\lt360°$ $θ'=θ=n·360°$, where $n$ is an integer. $θ'=495°-360°=135°$ $sin~495°=sin~135°=\frac{\sqrt 2}{2}$ $cos~495°=cos~135°=-\frac{\sqrt 2}{2}$ $tan~495°=tan~135°=-1$
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