Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 8 - Applications of Trigonometric Functions - 8.1 Right Triangle Trigonometry; Applications - 8.1 Assess Your Understanding - Page 519: 19

Answer

$0$

Work Step by Step

We know that for any acute angle $\theta$, $\sin(\theta)=\cos(90^{o}-\theta)$. Hence, $\sin(38^{o})-\cos(52^{o}) \\=\cos(90^{o}-38^{o})-\cos(52^{o}) \\=\cos(52^{o})-\cos(52^{o}) \\=0$
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