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Q (2)   Prove that 

\small (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x )\cos x = 0

Answers (1)

best_answer

We know that 
         sin3x=3\sin x - 4\sin^{3}x
           and
        cos3x=4\cos^{3}x - 3\cos x
 We use this in our problem
    \small (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x )\cos x
  =  (3\sin x - 4\sin^{3}x+ sin x) sinx  + (4\cos^{3}x - 3\cos x- cos x)cos x 
  =   (4sinx - 4\small \sin^{3}x)sinx + (4\small \cos^{3}x - 4cos x)cosx
  now take the 4sinx common from 1st term and  -4cosx from 2nd term
=  4\small \sin^{2}x(1 - \small \sin^{2}x)  - 4\small \cos^{2}x(1 - \small \cos^{2}x)
= 4\small \sin^{2}x\small \cos^{2}x - 4\small \cos^{2}x\small \sin^{2}x                                                                                            \small \because \ \ \ \cos^{2}x = 1 - \sin^2x\\ and\\ \sin^{2}x = 1 -\cos^{2}x
= 0 = R.H.S.

Posted by

seema garhwal

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