Question

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

Answer 1

R.H.S. $={(sinx-siny)}^2+{(cosx+cosy)}^2$

$={\sin }^{2}x+{\sin }^{2}y-2\sin x\cdot \sin y+{\cos }^{2}x+{\cos }^{2}y+2\cos x\cdot \cos y$

$={\sin }^{2}x+{\cos }^{2}x+{\sin }^{2}y+{\cos }^{2}y+2\cos x\cdot \cos y-2\sin x\cdot \sin y$

$=1+1+2\cos (x+y)$

$=2+2[1-2si{n}^2\left(\frac{x+y}{2}\right)]$

$=2+2-4si{n}^2\left(\frac{x+y}{2}\right)$

$=4-4si{n}^2\left(\frac{x+y}{2}\right)$

$=$ L.H.S.