Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.2 Trigonometric Functions: Unit Circle Approach - 5.2 Assess Your Understanding - Page 403: 109



Work Step by Step

Recall the identity: $(f \cdot g) (x)=f(x) \cdot g(x)$ Let us consider that $f(x)=\sin x $ and $g(x)=\cos x$ We know from the unit circle that $\sin (\dfrac{3 \pi}{4}) =\dfrac{\sqrt {2}}{2}$ and $\cos (\dfrac{3 \pi}{4})=\dfrac{-\sqrt 2}{2}$ Thus, we have: $(f \cdot g) (60^{\circ}) =\sin (\dfrac{3 \pi}{4}) \cdot \cos (\dfrac{3 \pi}{4}) \\=\dfrac{\sqrt {2}}{2} \cdot [\dfrac{-\sqrt 2}{2} ]\\=\dfrac{-1}{2}$
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