Question

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

Answer 1

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

$L.H.S$

$⟹{\cos }^{2}x+{\cos }^{2}y+2\cos x\cos y+{\sin }^{2}x+{\sin }^{2}y+2\sin x\sin y$

$⟹({\cos }^{2}x+{\sin }^{2}x)+({\sin }^{2}y+{\cos }^{2}y)+2\cos x\cos y+2\sin x\sin y$

$⟹1+1+2\cos x\cos y+2\sin x\sin y$

$⟹2+\cos (x-y)+\cos (x+y)+\cos (x-y)-\cos (x+y)$$⟹2+2\cos (x-y)$

$⟹2\{1+\cos (x-y\left)\right\}$

$⟹2\times 2\times {\cos }^2\frac{(x-y)}{2}$

$⟹{4\cos }^2\frac{(x-y)}{2}$