Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 6 - Section 6.3 - Trigonometric Functions of Angles - 6.3 Exercises - Page 499: 56

Answer

$\frac{3}{4}$ , 0

Work Step by Step

Given expression is- $\sin^{2}\theta$ , $\sin\theta^{2}$ Also given $\theta$ = $\frac{\pi}{3}$ $\sin^{2}\theta$ = $(\sin\theta )^{2}$ = $(\sin\frac{\pi}{3} )^{2}$ = $(\frac{\sqrt 3}{2})^{2}$ = $\frac{3}{4}$ $\sin(\theta)^{2}$ = $\sin(\frac{\pi}{3})^{2}$ = $\sin \frac{\pi^{2}}{9}$ = $\sin \frac{180\times180}{9}$ (Substituting $\pi = 180^{0}$) = $\sin 3600^{0}$ Reference angle of 3600 will be 0 Therefore $\sin 3600^{0}$ = $\sin 0^{0}$ = 0 i.e. $\sin(\frac{\pi}{3})^{2}$ = 0 Hence $(\sin\frac{\pi}{3} )^{2}$ = $\frac{3}{4}$ and $\sin(\frac{\pi}{3})^{2}$ = 0
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