Q (13) Prove the following

\small \cos^{2}2x - \cos^{2}6x = \sin4x\sin8x

Answers (1)
S safeer

As we know that 
 

a^{2}-b^{2} =(a-b)(a+b)

\therefore \cos^{2}2x -\cos^{2}6x = (\cos2x-\cos6x)(\cos2x+\cos6x)
Now
       \cos A - \cos B = -2\sin\left ( \frac{A+B}{2} \right )\sin\left ( \frac{A-B}{2} \right )\\ \\ \cos A + \cos B = 2\cos\left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )
By using these identities

cos2x - cos6x = -2sin(4x)sin(-2x) = 2sin4xsin2x                  ( \because sin(-x) = -sin x
                                                                                                    cos(-x) = cosx)
cos2x + cos 6x = 2cos4xcos(-2x) = 2cos4xcos2x

So our equation becomes 


                                                                       R.H.S.

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