Question

If $x+y=\pi +z$, then prove that ${\sin }^{2}x+{\sin }^{2}y-{\sin }^{2}z=2\sin x\sin y\cos z$.

Answer 1

$x+y=\pi +z$

$\Rightarrow$$z=x+y-\pi$

$\Rightarrow {\sin }^{2}x+{\sin }^{2}y-{\sin }^{2}z$

$={\sin }^{2}x+{\sin }^{2}y-{\sin }^2(x+y-\pi )$

$={\sin }^{2}x+{\sin }^{2}y-{\sin }^2(x+y)$

$={\sin }^{2}x+\sin (-x)\times \sin (x+2y)$

$=\sin x\times (\sin x-\sin (x+2y\left)\right)$

$=\sin x\times 2\times \cos (x+y)\times \sin (-y)$

$=2\sin x$ $\sin y$ $\cos z$

L.H.S.= R.H.S.