cos ^3left(frac{pi}{8}right) cos left(frac{3 | vedclass

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(A) Let the given expression be $E = \cos ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{3 \pi}{8}\right)+\sin ^3\left(\frac{\pi}{8}\right) \sin \left(

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$\frac{1}{2 \sqrt{2}}$

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(A) Let the given expression be $E = \cos ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{3 \pi}{8}\right)+\sin ^3\left(\frac{\pi}{8}\right) \sin \left(\frac{3 \pi}{8}\right)$.Using the identity $\cos \left(\frac{3 \pi}{8}\right) = \sin \left(\frac{\pi}{2} - \frac{3 \pi}{8}\right) = \sin \left(\frac{\pi}{8}\right)$ and $\sin \left(\frac{3 \pi}{8}\right) = \cos \left(\frac{\pi}{2} - \frac{3 \pi}{8}\right) = \cos \left(\frac{\pi}{8}\right)$,we substitute these into the expression:$E = \cos ^3\left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right) + \sin ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$Factor out $\sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$:$E = \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) \left[ \cos ^2\left(\frac{\pi}{8}\right) + \sin ^2\left(\frac{\pi}{8}\right) \right]$Using the Pythagorean identity $\sin ^2 A + \cos ^2 A = 1$:$E = \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) \cdot (1) = \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$Multiply and divide by $2$ to use the double angle formula $\sin (2A) = 2 \sin A \cos A$:$E = \frac{1}{2} \left[ 2 \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) \right] = \frac{1}{2} \sin \left( 2 \cdot \frac{\pi}{8} \right) = \frac{1}{2} \sin \left(\frac{\pi}{4}\right)$Since $\sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$:$E = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2 \sqrt{2}}$.

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

$\cos ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{3 \pi}{8}\right)+\sin ^3\left(\frac{\pi}{8}\right) \sin \left(\frac{3 \pi}{8}\right) = $

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