cos ^4 frac{pi}{12} + cos ^4 frac{5 pi}{12} + | vedclass

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(C) Let $	heta_1 = \frac{\pi}{12}$,$	heta_2 = \frac{5\pi}{12}$,$	heta_3 = \frac{7\pi}{12}$,and $	heta_4 = \frac{11\pi}{12}$.Note that $	heta_3 =

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$\frac{7}{4}$

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(C) Let $	heta_1 = \frac{\pi}{12}$,$	heta_2 = \frac{5\pi}{12}$,$	heta_3 = \frac{7\pi}{12}$,and $	heta_4 = \frac{11\pi}{12}$.Note that $	heta_3 = \pi - 	heta_2$ and $	heta_4 = \pi - 	heta_1$.Since $\cos(\pi - x) = -\cos x$,we have $\cos^4(\pi - x) = (-\cos x)^4 = \cos^4 x$.Thus,the expression becomes $2(\cos^4 \frac{\pi}{12} + \cos^4 \frac{5\pi}{12})$.Using $\cos^2 x = \frac{1 + \cos 2x}{2}$,we have $\cos^4 x = \frac{(1 + \cos 2x)^2}{4} = \frac{1 + 2\cos 2x + \cos^2 2x}{4} = \frac{1 + 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{3 + 4\cos 2x + \cos 4x}{8}$.For $x = \frac{\pi}{12}$,$2x = \frac{\pi}{6}$ and $4x = \frac{\pi}{3}$. So $\cos^4 \frac{\pi}{12} = \frac{3 + 4(\frac{\sqrt{3}}{2}) + \frac{1}{2}}{8} = \frac{3.5 + 2\sqrt{3}}{8}$.For $x = \frac{5\pi}{12}$,$2x = \frac{5\pi}{6}$ and $4x = \frac{5\pi}{3}$. So $\cos^4 \frac{5\pi}{12} = \frac{3 + 4(-\frac{\sqrt{3}}{2}) + \frac{1}{2}}{8} = \frac{3.5 - 2\sqrt{3}}{8}$.Summing these: $2 	imes (\frac{3.5 + 2\sqrt{3} + 3.5 - 2\sqrt{3}}{8}) = 2 	imes \frac{7}{8} = \frac{7}{4}$.

$\cos ^4 \frac{\pi}{12} + \cos ^4 \frac{5 \pi}{12} + \cos ^4 \frac{7 \pi}{12} + \cos ^4 \frac{11 \pi}{12} = $

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