Question

Find the value of,
$\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)$
by using formula,
$\cos A\cos B-\sin A\sin B=\cos (A+B)$

Answer 1

$\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)\to \left(1\right)$

as given

$\cos (A+B)=\cos A\cos B-\sin A\sin B$

so here $A=\frac{\pi }{4}-x,B=\frac{\pi }{4}-y$

so equation $\left(1\right)$ equals to

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

$\Rightarrow \cos (\frac{\pi }{2}-x-y)$

$\Rightarrow \cos (\frac{\pi }{2}-(x+y))$

$=\sin (x+y)$