## Calculus 8th Edition

(a) $sinx$ $cosy$ $=\frac{1}{2}[sin(x+y)+sin(x-y)]$ (b) $cosx$ $cosy$ $=\frac{1}{2}[cos(x+y)+cos(x-y)]$ (c) $sinx$ $siny$ $=\frac{1}{2}[cos(x-y)-cos(x+y)]$
(a) Since, $sin(x+y)=sinxcosy+cosxsiny$ and $sin(x-y)=sinxcosy-cosxsiny$ Thus, $sin(x+y)+sin(x-y)=(sinxcosy+cosxsiny)+(cosxcosy-sinxsiny)$ $sin(x+y)+sin(x-y)=(sinxcosy+cosxsiny)+(sinxcosy-cosxsiny)$ $sin(x+y)+sin(x-y)=(2sinxcosy)$ Hence, $sinx$ $cosy$ $=\frac{1}{2}[sin(x+y)+sin(x-y)]$ (b) Since, $cos(x+y)=cosxcosy-sinxsiny$ and $cos(x-y)=cosxcosy+sinxsiny$ Thus, $cos(x+y)+cos(x-y)=(cosxcosy-sinxsiny)+(cosxcosy+sinxsiny)$ $cos(x+y)+cos(x-y)=(2cosxcosy)$ Hence, $cosx$ $cosy$ $=\frac{1}{2}[cos(x+y)+cos(x-y)]$ (c) Since, $cos(x+y)=cosxcosy-sinxsiny$ and $cos(x-y)=cosxcosy+sinxsiny$ Thus, $cos(x-y)-cos(x+y)=(cosxcosy+sinxsiny)-(cosxcosy-sinxsiny)$ $cos(x-y)-cos(x+y)=(2sinx siny)$ Hence, $sinx$ $siny$ $=\frac{1}{2}[cos(x-y)-cos(x+y)]$