Question

simplify the given equation $\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)=\sin (x+y)$

Answer 1

$\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)=\sin (x+y)$
we know that $\cos (A-B)=\cos A\cos B-\sin A\sin B$
$\cos ((\frac{z}{4}-x)+(\frac{z}{4}-y))$
$\cos (\frac{z}{2}-(x+y))=\sin (x+y)$
As $\cos (90-\theta )=\sin \theta$