Q(7)   Prove that 

\small \sin3x + \sin2x - \sin x = 4\sin x \cos\frac{x}{2}\cos\frac{3x}{2}

Answers (1)

We know that 
 cosA + cosB = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
 sinA - sinB = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}
           
we use these identities
                                      sin3x - sinx = 2\cos\frac{3x+x}{2}\sin\frac{3x-x}{2} 

                                                                        = 2\cos2x\sin x

 
       sin2x + 2\cos2x\sin x  =   2sinx cosx  + 2\cos2x\sin x
  take 2 sinx common 
                       2sinx ( cosx + cos2x) = 2sinx(2\cos\frac{2x+x}{2}\cos\frac{2x-x}{2})
             
                                                            = 2sinx(2\cos\frac{3x}{2}\cos\frac{x}{2})
                                                           = 4sinx\cos\frac{3x}{2}\cos\frac{x}{2}

                                                               =  R.H.S.

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