Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - Exercise Set - Page 626: 48



Work Step by Step

Suppose $\theta =\tan^{-1} (\dfrac{7}{24})$ This gives: $\tan \theta=\dfrac{7}{24}$ Since, $ r=\sqrt {x^2+y^2}$ This implies that $ x=\sqrt {r^2-y^2}=\sqrt {(24)^2+(7)^2}=25$ Therefore, we have $\sin \theta =\sin [\tan^{-1} (\dfrac{7}{24})]=\dfrac{7}{25}$
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