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Q(5)  Prove that 

\small \sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos2x \sin4x

Answers (1)


we know that 
                     sinA + sinB =2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
We use this identity in our problem
If we notice we need sin4x in our final result so it is better if we made a combination of sin7x and sin x , sin3x and  sin5x tp get sin4x

 (sin7x + sinx) + (sin5x + sin3x) = 2\sin\frac{7x+x}{2}\cos\frac{7x-x}{2} +2\sin\frac{5x+3x}{2}\cos\frac{5x-3x}{2}
                                                                              =2\sin4x\cos3x + 2\sin4x\cos x
take 2sin4x common
                                                       = 2sin4x(cos3x + cosx)
We know that 
                     cosA + cosB =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
We use this
                  cos3x + cosx =2\cos\frac{3x+x}{2}\cos\frac{3x-x}{2}
                                          = 2\cos2x\cos x
                                          = 2sin4x(2\cos2x\cos x)
                                          = 4cosxcos2xsin4x = R.H.S.

Posted by

seema garhwal

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