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Question: \[(\cos\alpha + \cos\beta)^{2} + (\sin\alpha + \sin\beta)^{2} =\]...

(cos⁡α+cos⁡β)2+(sin⁡α+sin⁡β)2=(\cos\alpha + \cos\beta)^{2} + (\sin\alpha + \sin\beta)^{2} =

A

4cos⁡2α−β24\cos^{2}\frac{\alpha - \beta}{2}

B

4sin⁡2α−β24\sin^{2}\frac{\alpha - \beta}{2}

C

4cos⁡2α+β24\cos^{2}\frac{\alpha + \beta}{2}

D

4sin⁡2α+β24\sin^{2}\frac{\alpha + \beta}{2}

Answer

4cos⁡2α−β24\cos^{2}\frac{\alpha - \beta}{2}

Explanation

Solution

(cos⁡α+cos⁡β)2+(sin⁡α+sin⁡β)2(\cos\alpha + \cos\beta)^{2} + (\sin\alpha + \sin\beta)^{2}

=cos⁡2α+cos⁡2β+2cos⁡αcos⁡β+sin⁡2α+= \cos ^ { 2 } \alpha + \cos ^ { 2 } \beta + 2 \cos \alpha \cos \beta + \sin ^ { 2 } \alpha + sin⁡2β+2sin⁡αsin⁡β\sin^{2}\beta + 2\sin\alpha\sin\beta

=2{1+cos⁡(α−β)}=4cos⁡2(α−β2)= 2\{ 1 + \cos(\alpha - \beta)\} = 4\cos^{2}\left( \frac{\alpha - \beta}{2} \right).