Algebra and Trigonometry 10th Edition

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

Chapter 7 - 7.4 - Sum and Difference Equations - 7.4 Exercises - Page 539: 57


The identity is verified. $sin(\frac{\pi}{2}-x)=cos~x$

Work Step by Step

We know that: $sin\frac{\pi}{2}=1$ and $cos\frac{\pi}{2}=0$ $sin(u-v)=sin~u~cos~v-cos~u~sin~v$ $sin(\frac{\pi}{2}-x)=sin~\frac{\pi}{2}~cos~x-cos~\frac{\pi}{2}~sin~x=1~cos~x-0~sin~x=cos~x$
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