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)

Collected in the board: Trigonometry

Steven Zheng posted 18 hours ago

Answer

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)


Steven Zheng posted 18 hours ago

Scroll to Top