## Calculus: Early Transcendentals 8th Edition

$sin(x+y)$ $sin(x-y)$ $=sin^{2}x-sin^{2}y$
We need to prove the identity $sin(x+y)$ $sin(x-y)$ $=sin^{2}x-sin^{2}y$ Let us solve the 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, $sin(x+y)$ $sin(x-y)$ $=sin^{2}x-sin^{2}y$