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

$L.H.S=\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)$
Using formula, $\cos (A+B)=\cos A\cos B-\sin A\sin B$
Let $A=\frac{\pi }{4}-x$ and $B=\frac{\pi }{4}-y$
$\therefore \cos [(\frac{\pi }{4}-x)+(\frac{\pi }{4}-y)]$
$=\cos [\frac{\pi }{4}-x+\frac{\pi }{4}-y]$
$=\cos [\frac{\pi }{2}-(x+y\left)\right]=\sin (x+y)\{\because \cos (\frac{\pi }{2}-\theta )=\sin \theta \}$
So, LH.S =R.H.S
$\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)=\sin (x+y)$
Hence proved.