Answer
$sin(x+y)$ $sin(x-y)$ $=sin^{2}x-sin^{2}y$
Work Step by Step
Need to prove the identity
$sin(x+y)$ $sin(x-y)$ $=sin^{2}x-sin^{2}y$
Let us solve left side of the given identity.
$sin(x+y)$ $sin(x-y)$ $=(sinxcosy+cosxsiny)\times (sinxcosy-cosxsiny)$
$sin(x+y)$ $sin(x-y)$ $=sin^{2}xcos^{2}y-cos^{2}xsin^{2}y$
$sin(x+y)$ $sin(x-y)$ $=sin^{2}x(1-sin^{2}y)-(1-sin^{2}x)sin^{2}y$
$sin(x+y)$ $sin(x-y)$ $=(sin^{2}x-sin^{2}xsin^{2}y)-(sin^{2}y-sin^{2}xsin^{2}y)$
$sin(x+y)$ $sin(x-y)$ $=sin^{2}x-sin^{2}xsin^{2}y-sin^{2}y+sin^{2}xsin^{2}y$
Hence, $sin(x+y)$ $sin(x-y)$ $=sin^{2}x-sin^{2}y$
