Question

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

Answer 1

Hint: Note that, $\cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right)$
Again, $\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta $
Therefore take the left hand side and use these two formulae to proceed.
And on solving we will arrive at our desired result.

Complete step-by-step answer:
Given to prove $\cos \left( {\dfrac{\pi }{4} - x} \right)\cos \left( {\dfrac{\pi }{4} - y} \right) - \sin \left( {\dfrac{\pi }{4} - x} \right)\sin \left( {\dfrac{\pi }{4} - y} \right) = \sin \left( {x + y} \right)$,
We know that, $\cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right)$
Now, left hand side is given by,
$ = \cos \left( {\dfrac{\pi }{4} - x} \right)\cos \left( {\dfrac{\pi }{4} - y} \right) - \sin \left( {\dfrac{\pi }{4} - x} \right)\sin \left( {\dfrac{\pi }{4} - y} \right)$
Using, $\cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right)$, we get,
$ = \cos \left( {\dfrac{\pi }{4} - x + \dfrac{\pi }{4} - y} \right)$
On simplification we get,
$ = \cos \left( {\dfrac{\pi }{2} - \left( {x + y} \right)} \right)$
Using, $\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta $, we get,
$ = \sin \left( {x + y} \right)$
= Right hand side
Therefore, $\cos \left( {\dfrac{\pi }{4} - x} \right)\cos \left( {\dfrac{\pi }{4} - y} \right) - \sin \left( {\dfrac{\pi }{4} - x} \right)\sin \left( {\dfrac{\pi }{4} - y} \right) = \sin \left( {x + y} \right)$ (proved).

Note: Note the following important formulae:
1.$\sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x{\text{ , tan}}\left( {\dfrac{\pi }{2} - x} \right) = 2.\cot x{\text{ , cosec}}\left( {\dfrac{\pi }{2} - x} \right) = \sec x$
3.$\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)$
4.$\sin A\cos B - \cos A\sin B = \sin \left( {A - B} \right)$
5.$\cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right)$
6.$\cos A\cos B + \sin A\sin B = \cos \left( {A - B} \right)$
7.$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
8.$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$