Question

Prove the following:

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

Answer 1

We have

L.H.S = cos $(\frac{\pi }{4}-x)$ cos $(\frac{\pi }{4}-y)$

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

[$\because$ cos (A+B) = cos A cos B- sin A sin B]

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

= sin (x+y) = R.H.S.