sin ^4 frac{pi}{8}+cos ^4 frac{pi}{8}+sin ^4 | vedclass

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Question

(A) We know that $\sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2 - 2 \sin^2 	heta \cos^2 	heta = 1 - \frac{1}{2} \sin^2(2 	heta)

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) We know that $\sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2 - 2 \sin^2 	heta \cos^2 	heta = 1 - \frac{1}{2} \sin^2(2 	heta)$.Let $S = \sum_{k=1,3,5,7} (\sin^4 \frac{k \pi}{8} + \cos^4 \frac{k \pi}{8})$.Using the identity,$S = \sum_{k=1,3,5,7} (1 - \frac{1}{2} \sin^2 \frac{k \pi}{4})$.Since there are $4$ terms,$S = 4 - \frac{1}{2} (\sin^2 \frac{\pi}{4} + \sin^2 \frac{3 \pi}{4} + \sin^2 \frac{5 \pi}{4} + \sin^2 \frac{7 \pi}{4})$.We know $\sin^2 \frac{\pi}{4} = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$,$\sin^2 \frac{3 \pi}{4} = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$,$\sin^2 \frac{5 \pi}{4} = (-\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$,and $\sin^2 \frac{7 \pi}{4} = (-\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.Thus,$S = 4 - \frac{1}{2} (\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}) = 4 - \frac{1}{2} (2) = 4 - 1 = 3$.

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

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