14,sin2 x + 2 sin 4x + sin 6r = 4 cos2 x sin 4x

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Solution

L.H.S = sin2x+2sin4x+sin6x

Use trigonometric identity

sinA+sinB=2sin( A+B 2 )⋅cos( A−B 2 )

Simplifying the L.H.S,

sin2x+2sin4x+sin6x =[ 2sin( 2x+6x 2 )⋅cos( 2x−6x 2 ) ]+2sin4x =2sin4x⋅cos( −2x )+2sin4x =2sin4x⋅cos2x+2sin4x =2sin4x( cos2x+1 )

=2sin4x⋅( 2 cos 2 x−1+1 ) [∵cos2θ = 2cos 2 θ - 1] =2sin4x⋅( 2 cos 2 x ) =4 cos 2 x⋅sin4x

L.H.S. = R.H.S.

Hence, proved.

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