Math, asked by PragyaTbia, 11 months ago

निम्नलिखित को सिद्ध कीजिए: \sin^2{x} + 2 \sin 4x + \sin 6x = 4 \cos^2{x}\,\sin 4x

Answers

Answered by kaushalinspire
0

Answer:

Step-by-step explanation:

sin2x + 2 sin4x + sin6x = 4 cos^{2}x sin4x

L.H.S.  =  sin2x + 2 sin4x + sin6x

          =  ( sin2x  + sin6x )+ 2 sin4x

          =  2 sin( \frac{2x+6x}{2} ) [tex]cos\frac{2x-6x}{2} + 2 sin4x

         (∵ sinA + sinB = 2sin\frac{A+B}{2} cos\frac{A-B}{2} )

         =   2 sin(4x) cos(-2x)  + 2 sin4x

        =   2 sin4x ( cos2x + 1 )         [∵cos(-θ) = cosθ ]

        =  2 sin4x ( 2 cos^{2}x )         [ 1+cos2x = 2cos^{2}x]

        =  4 cos^{2}x sin4x

         = R.H.S.

        L.H.S. = R.H.S.

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