If theta is an acute angle and 2 sin ^2 theta | vedclass

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(C) Given the expression: $2 \sin ^2 	heta = \cos ^4 \frac{\pi}{8} + \sin ^4 \frac{3 \pi}{8} + \cos ^4 \frac{5 \pi}{8} + \sin ^4 \frac{7 \pi}{8}$ Usi

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$\frac{\pi}{3}$

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(C) Given the expression: $2 \sin ^2 	heta = \cos ^4 \frac{\pi}{8} + \sin ^4 \frac{3 \pi}{8} + \cos ^4 \frac{5 \pi}{8} + \sin ^4 \frac{7 \pi}{8}$ Using the identities $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we have: $\sin^4(\frac{7\pi}{8}) = \sin^4(\pi - \frac{\pi}{8}) = \sin^4(\frac{\pi}{8})$ $\cos^4(\frac{5\pi}{8}) = \cos^4(\pi - \frac{3\pi}{8}) = (-\cos(\frac{3\pi}{8}))^4 = \cos^4(\frac{3\pi}{8})$ Substituting these into the expression: $2 \sin ^2 	heta = (\cos ^4 \frac{\pi}{8} + \sin ^4 \frac{\pi}{8}) + (\sin ^4 \frac{3 \pi}{8} + \cos ^4 \frac{3 \pi}{8})$ Using $a^2 + b^2 = (a+b)^2 - 2ab$: $= [(\cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8})^2 - 2 \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}] + [(\sin^2 \frac{3\pi}{8} + \cos^2 \frac{3\pi}{8})^2 - 2 \sin^2 \frac{3\pi}{8} \cos^2 \frac{3\pi}{8}]$ $= [1 - \frac{1}{2} \sin^2 \frac{\pi}{4}] + [1 - \frac{1}{2} \sin^2 \frac{3\pi}{4}]$ $= 2 - \frac{1}{2} [(\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2] = 2 - \frac{1}{2} [\frac{1}{2} + \frac{1}{2}] = 2 - \frac{1}{2} = \frac{3}{2}$ Thus,$2 \sin^2 	heta = \frac{3}{2} \implies \sin^2 	heta = \frac{3}{4} \implies \sin 	heta = \frac{\sqrt{3}}{2}$ Since $	heta$ is an acute angle,$	heta = \frac{\pi}{3}$.

If $\theta$ is an acute angle and $2 \sin ^2 \theta = \cos ^4 \frac{\pi}{8} + \sin ^4 \frac{3 \pi}{8} + \cos ^4 \frac{5 \pi}{8} + \sin ^4 \frac{7 \pi}{8}$,then $\theta =$

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