Question

Evaluate

$\begin{array}{cc}& {\cos }^6\frac{\pi }{16}+{\cos }^6\frac{3\pi }{16}+{\cos }^6\frac{5\pi }{16}+{\cos }^6\frac{7\pi }{16}\\ & \end{array}$

Answer 1

$\cos {}^6\frac{\pi }{16}+\cos {}^6\frac{3\pi }{16}+\cos {}^6\frac{5\pi }{16}+\cos {}^6\frac{7\pi }{16}=\cos {}^6\frac{\pi }{16}+\cos {}^6(\frac{\pi }{2}-\frac{\pi }{16})+\cos {}^6\frac{3\pi }{16}+\cos {}^6(\frac{\pi }{2}-\frac{3\pi }{16})=(\cos {}^6\frac{\pi }{16}+\sin {}^6\frac{\pi }{16})+(\cos {}^6\frac{3\pi }{16}+\sin {}^6\frac{3\pi }{16})=\left[\right(\cos {}^2\frac{\pi }{16}+\sin {}^2\frac{\pi }{16}\left)\right(\cos {}^4\frac{\pi }{16}-\cos {}^2\frac{\pi }{16}\sin {}^2\frac{\pi }{16}+\sin {}^2\frac{\pi }{16}\left)\right]+\left[\right(\cos {}^2\frac{3\pi }{16}+\sin {}^2\frac{3\pi }{16}\left)\right(\cos {}^4\frac{3\pi }{16}-\cos {}^4\frac{3\pi }{16}\sin {}^2\frac{3\pi }{16}+\sin {}^2\frac{3\pi }{16}\left)\right]=[\cos {}^4\frac{\pi }{16}-\cos {}^2\frac{\pi }{16}\sin {}^2\frac{\pi }{16}+\sin {}^2\frac{\pi }{16}]+[\cos {}^4\frac{3\pi }{16}-\cos {}^2\frac{3\pi }{16}\sin {}^2\frac{3\pi }{16}+\sin {}^2\frac{3\pi }{16}]=[{[\cos {}^2\frac{\pi }{16}+\sin {}^2\frac{\pi }{16}]}^2-3\cos {}^2\frac{\pi }{16}\sin {}^2\frac{\pi }{16}]+[{[\cos {}^2\frac{3\pi }{16}+\sin {}^2\frac{3\pi }{16}]}^2-3\cos {}^2\frac{3\pi }{16}\sin {}^2\frac{3\pi }{16}]=2-3\cos {}^2\frac{\pi }{16}\sin {}^2\frac{\pi }{16}-3\cos {}^2\frac{3\pi }{16}\sin {}^2\frac{3\pi }{16}=2-\frac{3}{4}{\left(2\sin \frac{\pi }{16}\cos \frac{\pi }{16}\right)}^2-\frac{3}{4}{\left(2\sin \frac{3\pi }{16}\cos \frac{3\pi }{16}\right)}^2=2-\frac{3}{4}[\sin {}^2\frac{\pi }{8}+\sin {}^2\frac{3\pi }{8}]=2-\frac{3}{4}[\sin {}^2\frac{\pi }{8}+\sin {}^2[\frac{\pi }{2}-\frac{\pi }{8}]]=2-\frac{3}{4}[\sin {}^2\frac{\pi }{8}+\cos {}^2\frac{\pi }{8}]=2-\frac{3}{4}=\frac{5}{4}$