Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - P.S. Problem Solving - Page 557: 3


The identity is verified. $sin[\frac{(12n+1)\pi}{6}]=\frac{1}{2}$ for all integers $n$

Work Step by Step

Use: $sin(a+b)=sin~a~cos~b+cos~a~sin~b$ $sin[\frac{(12n+1)\pi}{6}]=sin[(2\pi)n+\frac{\pi}{6}]=sin~(2\pi)n~cos\frac{\pi}{6}+cos~(2\pi)n~sin\frac{\pi}{6}$ But, $sin~(2\pi)n=0$ for all integers $n$ $cos~(2\pi)n=1$ for all integers $n$ $sin[\frac{(12n+1)\pi}{6}]=sin~(2\pi)n~cos\frac{\pi}{6}+cos~(2\pi)n~sin\frac{\pi}{6}=0~(\frac{\sqrt 3}{2})+1~(\frac{1}{2})=\frac{1}{2}$
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