University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 28: 43

Answer

$$\sin\frac{7\pi}{12}=\frac{\sqrt2+\sqrt6}{4}$$

Work Step by Step

As suggested by the exercise, we would rewrite: $$\frac{7\pi}{12}=\frac{\pi}{4}+\frac{\pi}{3}$$ Therefore, $$\sin\frac{7\pi}{12}=\sin\Big(\frac{\pi}{4}+\frac{\pi}{3}\Big)$$ Apply the addition formula for sine here: $$\sin\frac{7\pi}{12}=\sin\frac{\pi}{4}\cos\frac{\pi}{3}+\cos\frac{\pi}{4}\sin\frac{\pi}{3}$$ Remember that $\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt2}{2}$ and $\sin\frac{\pi}{3}=\frac{\sqrt3}{2}$ and $\cos\frac{\pi}{3}=\frac{1}{2}$ $$\sin\frac{7\pi}{12}=\frac{\sqrt2}{2}\times\frac{1}{2}+\frac{\sqrt2}{2}\times\frac{\sqrt3}{2}$$ $$\sin\frac{7\pi}{12}=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}$$ $$\sin\frac{7\pi}{12}=\frac{\sqrt2+\sqrt6}{4}$$
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