(cos alpha + cos beta )^2 + (sin alpha + sin | vedclass

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Question

(A) Expanding the expression: $(\cos \alpha + \cos \beta )^2 + (\sin \alpha + \sin \beta )^2 = \cos ^2 \alpha + \cos ^2 \beta + 2\cos \alpha \cos \bet

Vedclass · Accepted Answer

$4\cos ^2\frac{\alpha - \beta }{2}$

Vedclass · Answer

(A) Expanding the expression: $(\cos \alpha + \cos \beta )^2 + (\sin \alpha + \sin \beta )^2 = \cos ^2 \alpha + \cos ^2 \beta + 2\cos \alpha \cos \beta + \sin ^2 \alpha + \sin ^2 \beta + 2\sin \alpha \sin \beta$ Using the identity $\sin ^2 	heta + \cos ^2 	heta = 1$: $= (\cos ^2 \alpha + \sin ^2 \alpha) + (\cos ^2 \beta + \sin ^2 \beta) + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta)$ $= 1 + 1 + 2\cos (\alpha - \beta)$ $= 2 + 2\cos (\alpha - \beta)$ $= 2(1 + \cos (\alpha - \beta))$ Using the identity $1 + \cos 	heta = 2\cos ^2 \frac{	heta}{2}$: $= 2 	imes 2\cos ^2 \left( \frac{\alpha - \beta }{2} ight)$ $= 4\cos ^2 \left( \frac{\alpha - \beta }{2} ight)$.

$(\cos \alpha + \cos \beta )^2 + (\sin \alpha + \sin \beta )^2 = $

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