cos 2(theta + phi ) - 4cos (theta + phi )sin | 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 \

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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$$E = \cos 2(	heta + \phi) - 2\cos(	heta + \phi)\cos(	heta - \phi) + 2\cos^2(	heta + \phi) + 2\sin^2 \phi$Using $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:$E = \cos 2(	heta + \phi) - [\cos 2	heta + \cos 2\phi] + 2\cos^2(	heta + \phi) + 2\sin^2 \phi$Using $2\cos^2 A = 1 + \cos 2A$:$E = \cos 2(	heta + \phi) - \cos 2	heta - \cos 2\phi + 1 + \cos 2(	heta + \phi) + 2\sin^2 \phi$Since $1 - \cos 2\phi = 2\sin^2 \phi$,the expression simplifies to:$E = 2\cos 2(	heta + \phi) - \cos 2	heta + 2\sin^2 \phi + 2\sin^2 \phi - \cos 2\phi$Alternatively,testing with $	heta = \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}) = \cos \pi - 0 + 2(\frac{1}{2}) = -1 + 1 = 0$.Checking options for $	heta = \frac{\pi}{4}$:$(a) \cos 2(\frac{\pi}{4}) = \cos \frac{\pi}{2} = 0$.Thus,option $(a)$ is correct.

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

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