Prove that : ( x+ y)^2+( x- y)^2 = 4 ^2 x + y 2

Question

Prove that : $(\cos x+\cos y)^2+(\sin x-\sin y)^2 = 4{\cos ^2}\frac{{x + y}}{2}$

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Answer

We have L.H.S. $=(\cos x+\cos y)^2+(\sin x-\sin y)^2$
$ = {\left[ {2\cos \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right)} \right]^2}$$ + {\left[ {2\cos \left( {\frac{{x + y}}{2}} \right)\sin \left( {\frac{{x - y}}{2}} \right)} \right]^2}$
$ = 4{\cos ^2}\left( {\frac{{x + y}}{2}} \right){\cos ^2}\left( {\frac{{x - y}}{2}} \right)$$ + 4{\cos ^2}\left( {\frac{{x + y}}{2}} \right){\sin ^2}\left( {\frac{{x - y}}{2}} \right)$
$ = 4{\cos ^2}\left( {\frac{{x + y}}{2}} \right)$$\left[ {{{\cos }^2}\left( {\frac{{x - y}}{2}} \right) + {{\sin }^2}\left( {\frac{{x - y}}{2}} \right)} \right]$
$ = 4{\cos ^2}\left( {\frac{{x + y}}{2}} \right) = R.H.S.$

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