Question

Prove that, $cos(\frac{\pi}{4}-x)cos(\frac{\pi}{4}-y)-sin(\frac{\pi}{4}-x)sin(\frac{\pi}{4}-y)=sin(x+y)$

Answer 1

$cos(\frac{\pi}{4}-x)cos(\frac{\pi}{4}-y)-sin(\frac{\pi}{4}-x)sin(\frac{\pi}{4}-y)$

$=\frac{1}{2}[2\,cos(\frac{\pi}{4}-x)cos(\frac{\pi}{4}-y)+\frac{1}{2}[-2sin(\frac{\pi}{4}-x)sin(\frac{\pi}{4}-y)]$

$=\frac{1}{2}[cos\\{(\frac{\pi}{4}-x)+(\frac{\pi}{4}-y)\\}+\\{cos(\frac{\pi}{4}-x)-(\frac{\pi}{4}-y)\\}]$

$+\frac{1}{2}[\,cos\\{(\frac{\pi}{4}-x)\\}+(\frac{\pi}{4}-y)-cos+\\{(\frac{\pi}{4}-x)-(\frac{\pi}{4}-y)\\}]$

$[∵ 2cos AcosB=cos(A+B)+cos(A-B)]$

$[-2sinAsinB=cos(A+B)-cos(A-B)]$

$=2×\frac{1}{2}[cos\\{(\frac{\pi}{4}-x)+(\frac{\pi}{4}-y)\\}]$

$=cos[\frac{\pi}{2}-(x+y)]$

$sin(x+y)$

$R.H.S$