## Algebra and Trigonometry 10th Edition

The identity is verified. $sin[\frac{(12n+1)\pi}{6}]=\frac{1}{2}$ for all integers $n$
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 $n$ $sin[\frac{(12n+1)\pi}{6}]=sin~(2\pi)n~cos\frac{\pi}{6}+cos~(2\pi)n~sin\frac{\pi}{6}=0~(\frac{\sqrt 3}{2})+1~(\frac{1}{2})=\frac{1}{2}$