cos 2(theta + phi) - 4cos (theta + phi)sin th | vedclass

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Question

(A) Given expression: $E = \cos 2(	heta + \phi) - 4\cos (	heta + \phi)\sin 	heta \sin \phi + 2\sin^2 \phi$Using the identity $2\sin 	heta \sin \ph

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$\cos 2	heta$

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(A) Given expression: $E = \cos 2(	heta + \phi) - 4\cos (	heta + \phi)\sin 	heta \sin \phi + 2\sin^2 \phi$Using the identity $2\sin 	heta \sin \phi = \cos(	heta - \phi) - \cos(	heta + \phi)$:$E = \cos 2(	heta + \phi) - 2\cos(	heta + \phi)[\cos(	heta - \phi) - \cos(	heta + \phi)] + 2\sin^2 \phi$Using $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:$E = \cos 2(	heta + \phi) - 2[\cos 2	heta + \cos 2\phi] + 2\cos 2(	heta + \phi) + 2\sin^2 \phi$Alternatively,using the substitution method:Let $	heta = \frac{\pi}{4}$ and $\phi = \frac{\pi}{4}$.$E = \cos 2(\frac{\pi}{2}) - 4\cos(\frac{\pi}{2})\sin(\frac{\pi}{4})\sin(\frac{\pi}{4}) + 2\sin^2(\frac{\pi}{4})$$E = \cos \pi - 4(0)(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + 2(\frac{1}{2}) = -1 - 0 + 1 = 0$.Checking options for $	heta = \frac{\pi}{4}, \phi = \frac{\pi}{4}$:$(a) \cos 2	heta = \cos(\frac{\pi}{2}) = 0$.Since the expression matches option $(a)$,the correct answer is $A$.

$\cos 2(\theta + \phi) - 4\cos (\theta + \phi)\sin \theta \sin \phi + 2\sin^2 \phi = $

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