sin ^4 frac{pi}{8} + sin ^4 frac{3pi}{8} + si | vedclass

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Question

(C) We know that $\sin \frac{5\pi}{8} = \sin(\pi - \frac{3\pi}{8}) = \sin \frac{3\pi}{8}$ and $\sin \frac{7\pi}{8} = \sin(\pi - \frac{\pi}{8}) = \sin

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) We know that $\sin \frac{5\pi}{8} = \sin(\pi - \frac{3\pi}{8}) = \sin \frac{3\pi}{8}$ and $\sin \frac{7\pi}{8} = \sin(\pi - \frac{\pi}{8}) = \sin \frac{\pi}{8}$.Thus,the expression becomes $2(\sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8})$.Using the identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we have $\sin^4 	heta = (\frac{1 - \cos 2	heta}{2})^2 = \frac{1}{4}(1 - 2\cos 2	heta + \cos^2 2	heta) = \frac{1}{4}(1 - 2\cos 2	heta + \frac{1 + \cos 4	heta}{2}) = \frac{1}{8}(3 - 4\cos 2	heta + \cos 4	heta)$.For $	heta = \frac{\pi}{8}$: $\sin^4 \frac{\pi}{8} = \frac{1}{8}(3 - 4\cos \frac{\pi}{4} + \cos \frac{\pi}{2}) = \frac{1}{8}(3 - 4(\frac{1}{\sqrt{2}}) + 0) = \frac{3 - 2\sqrt{2}}{8}$.For $	heta = \frac{3\pi}{8}$: $\sin^4 \frac{3\pi}{8} = \frac{1}{8}(3 - 4\cos \frac{3\pi}{4} + \cos \frac{3\pi}{2}) = \frac{1}{8}(3 - 4(-\frac{1}{\sqrt{2}}) + 0) = \frac{3 + 2\sqrt{2}}{8}$.Summing these: $2(\frac{3 - 2\sqrt{2}}{8} + \frac{3 + 2\sqrt{2}}{8}) = 2(\frac{6}{8}) = 2(\frac{3}{4}) = \frac{3}{2}$.

$\sin ^4 \frac{\pi}{8} + \sin ^4 \frac{3\pi}{8} + \sin ^4 \frac{5\pi}{8} + \sin ^4 \frac{7\pi}{8} = $

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