Question

Prove the following:
$\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-x)=\sin (x+y)$

Answer 1

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

We know that, $\cos (A+B)=\cos A\cos B-\sin A\sin B$

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

$A+B=\frac{\pi }{2}-(x+y)$

$\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-x)=\cos (\frac{\pi }{2}-(x+y\left)\right)=\sin (x+y)$