The expression cos^2 phi + cos^2(theta + phi) | vedclass

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(B) Let $E = \cos^2 \phi + \cos^2(	heta + \phi) - 2 \cos 	heta \cos \phi \cos(	heta + \phi)$.Using the identity $\cos^2 A = \frac{1 + \cos 2A}{2}$,

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independent of $\phi$

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(B) Let $E = \cos^2 \phi + \cos^2(	heta + \phi) - 2 \cos 	heta \cos \phi \cos(	heta + \phi)$.Using the identity $\cos^2 A = \frac{1 + \cos 2A}{2}$, we have:$E = \frac{1 + \cos 2\phi}{2} + \frac{1 + \cos 2(	heta + \phi)}{2} - \cos 	heta [\cos(2\phi + 	heta) + \cos 	heta]$$E = 1 + \frac{1}{2} [\cos 2\phi + \cos(2\phi + 2	heta)] - \cos 	heta \cos(2\phi + 	heta) - \cos^2 	heta$Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:$E = 1 + \cos(2\phi + 	heta) \cos 	heta - \cos 	heta \cos(2\phi + 	heta) - \cos^2 	heta$$E = 1 - \cos^2 	heta = \sin^2 	heta$.Since the result is $\sin^2 	heta$, the expression is independent of $\phi$.

The expression $\cos^2 \phi + \cos^2(\theta + \phi) - 2 \cos \theta \cos \phi \cos(\theta + \phi)$ is

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