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Question: Prove the following: \(\cos \left( {\dfrac{\pi }{4} - x} \right)\cos \left( {\dfrac{\pi }{4} - y} ...

Prove the following:
cos⁡(π4−x)cos⁡(π4−y)−sin⁡(π4−x)sin⁡(π4−y)=sin⁡(x+y)\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)

Explanation

Solution

Note that, cos⁡Acos⁡B−sin⁡Asin⁡B=cos⁡(A+B)\cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right)
Again, cos⁡(π2−θ)=sin⁡θ\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⁡(π4−x)cos⁡(π4−y)−sin⁡(π4−x)sin⁡(π4−y)=sin⁡(x+y)\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⁡Acos⁡B−sin⁡Asin⁡B=cos⁡(A+B)\cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right)
Now, left hand side is given by,
=cos⁡(π4−x)cos⁡(π4−y)−sin⁡(π4−x)sin⁡(π4−y)= \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⁡Acos⁡B−sin⁡Asin⁡B=cos⁡(A+B)\cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right), we get,
=cos⁡(π4−x+π4−y)= \cos \left( {\dfrac{\pi }{4} - x + \dfrac{\pi }{4} - y} \right)
On simplification we get,
=cos⁡(π2−(x+y))= \cos \left( {\dfrac{\pi }{2} - \left( {x + y} \right)} \right)
Using, cos⁡(π2−θ)=sin⁡θ\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta , we get,
=sin⁡(x+y)= \sin \left( {x + y} \right)
= Right hand side
Therefore, cos⁡(π4−x)cos⁡(π4−y)−sin⁡(π4−x)sin⁡(π4−y)=sin⁡(x+y)\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⁡(π2−x)=cos⁡x , tan(π2−x)=2.cot⁡x , cosec(π2−x)=sec⁡x\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⁡Acos⁡B+cos⁡Asin⁡B=sin⁡(A+B)\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)
4.sin⁡Acos⁡B−cos⁡Asin⁡B=sin⁡(A−B)\sin A\cos B - \cos A\sin B = \sin \left( {A - B} \right)
5.cos⁡Acos⁡B−sin⁡Asin⁡B=cos⁡(A+B)\cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right)
6.cos⁡Acos⁡B+sin⁡Asin⁡B=cos⁡(A−B)\cos A\cos B + \sin A\sin B = \cos \left( {A - B} \right)
7.tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}
8.tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}