Question
Prove
\cos(\dfrac{\pi}{4}-x)\cos(\dfrac{\pi}{4}-y) -\sin(\dfrac{\pi}{4}-x)\sin(\dfrac{\pi}{4}-y) =\sin(x+y)
Prove
\cos(\dfrac{\pi}{4}-x)\cos(\dfrac{\pi}{4}-y) -\sin(\dfrac{\pi}{4}-x)\sin(\dfrac{\pi}{4}-y) =\sin(x+y)
Sum identity for cosine function
\cos(\alpha +\beta ) = \cos \alpha\cos \beta -\sin \alpha \sin \beta
Here
\alpha = \dfrac{\pi}{4}-x
\beta =\dfrac{\pi}{4}-y
Apply sum and cofunciton identities
\cos(\dfrac{\pi}{4}-x)\cos(\dfrac{\pi}{4}-y) -\sin(\dfrac{\pi}{4}-x)\sin(\dfrac{\pi}{4}-y)
= \cos(\dfrac{\pi}{2} -x-y)
=\sin(x+y)