Answer
(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) $sinx$ $siny$ $=\frac{1}{2}[cos(x-y)-cos(x+y)]$
Work Step by Step
(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, $sinx$ $siny$ $=\frac{1}{2}[cos(x-y)-cos(x+y)]$