Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Test - Page 97: 4c



Work Step by Step

$\cos 90^{o}=0$, Function Values of Special Angles: $\displaystyle \cos 60^{o}=\frac{1}{2},\qquad \cos 30^{o}=\frac{\sqrt{3}}{2}$ $\displaystyle \sin 30^{o}=\frac{1}{2}, \quad \sin 60^{o}=\frac{\sqrt{3}}{2}$ $LHS= \cos 90^{o}=0$ $RHS= \displaystyle \frac{1}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\cdot\frac{1}{2}=0=LHS$ The statement is true.
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