Answer
$\sin(x+y)\sin (x-y)=\sin^2 x-\sin^2 y$
Work Step by Step
Need to verify $\sin(x+y)\sin (x-y)=\sin^2 x-\sin^2 y$
This implies that
$\dfrac{1}{2}[\cos (x+y-(x-y))-\cos (x+y+(x-y))=\dfrac{1}{2}[\cos 2y-\cos 2x]$
or, $\dfrac{1}{2}[1-2\sin^2 y-(1-2\sin^2 x)]=\dfrac{1}{2}[1-2\sin^2 y-1+\sin^2 x)]$
or, $\dfrac{1}{2}[2\sin^2 y+2\sin^2 x]= \sin^2 x-\sin^2 y$ (RHS)
Hence, the left-hand side and right hand side are equal.
