Trig3.10

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Prove that \sin^4 \frac{\pi}{8}+\sin^4 \frac{3\pi}{8}+\sin^4 \frac{5\pi}{8}+\sin^4 \frac{7\pi}{8}=\frac{3}{2}\,

LHS=\frac{1}{4}[(1-\cos \frac{\pi}{4})^2+(1-\cos \frac{3\pi}{4})^2+(1-\cos \frac{5\pi}{4})^2+(1-\cos \frac{7\pi}{4})^2]\,

Now \cos \frac{3\pi}{4}=\cos (\pi-\frac{\pi}{4})=-\cos \frac{\pi}{4}, \cos \frac{5\pi}{4}=\cos (\pi+\frac{\pi}{4})=-\cos \frac{\pi}{4},\cos \frac{7\pi}{4}=\cos (2\pi-\frac{\pi}{4})=\cos \frac{\pi}{4}\,

Therefore,LHS=\frac{1}{4}[(1-\cos \frac{\pi}{4})^2+(1+\cos \frac{\pi}{4})^2+(1+\cos \frac{\pi}{4})^2+(1-\cos \frac{\pi}{4})^2]\,

\frac{1}{2}[(1-\cos \frac{\pi}{4})^2+(1+\cos \frac{\pi}{4})^2]\,

\frac{2}{2}[1+\cos^2 \frac{\pi}{4}]=1+\frac{1}{2}=\frac{3}{2}\, =RHS


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