The value of cos ^4 frac{pi}{8}+cos ^4 frac{2 | vedclass

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Question

(A) Let $S = \sum_{k=1}^{8} \cos^4 \frac{k\pi}{8}$.Using the property $\cos(\pi - 	heta) = -\cos 	heta$,we have $\cos^4(\pi - 	heta) = \cos^4 	het

Vedclass · Accepted Answer

$3$

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(A) Let $S = \sum_{k=1}^{8} \cos^4 \frac{k\pi}{8}$.Using the property $\cos(\pi - 	heta) = -\cos 	heta$,we have $\cos^4(\pi - 	heta) = \cos^4 	heta$.Thus,$\cos^4 \frac{7\pi}{8} = \cos^4 \frac{\pi}{8}$,$\cos^4 \frac{6\pi}{8} = \cos^4 \frac{2\pi}{8}$,and $\cos^4 \frac{5\pi}{8} = \cos^4 \frac{3\pi}{8}$.Also,$\cos \frac{4\pi}{8} = \cos \frac{\pi}{2} = 0$ and $\cos \frac{8\pi}{8} = \cos \pi = -1$.So,$S = 2(\cos^4 \frac{\pi}{8} + \cos^4 \frac{2\pi}{8} + \cos^4 \frac{3\pi}{8}) + 0^4 + (-1)^4$.$S = 2(\cos^4 \frac{\pi}{8} + \sin^4 \frac{\pi}{8} + (\frac{1}{\sqrt{2}})^4) + 1$.Using $\cos^4 	heta + \sin^4 	heta = 1 - 2\sin^2 	heta \cos^2 	heta = 1 - \frac{1}{2}\sin^2(2	heta)$:$S = 2(1 - \frac{1}{2}\sin^2 \frac{\pi}{4} + \frac{1}{4}) + 1 = 2(1 - \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4}) + 1 = 2(1 - \frac{1}{4} + \frac{1}{4}) + 1 = 2(1) + 1 = 3$.

The value of $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{2 \pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{4 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{6 \pi}{8}+\cos ^4 \frac{7 \pi}{8}+\cos ^4 \frac{8 \pi}{8}$ is equal to:

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