Question

25.cos 6x = 32 cos®x-48cos4 x + 18 cos2x-1

Answer 1

Given: cos6x=32 cos 6 x−48 cos 4 x+18 cos 2 x−1

Consider the L.H.S of the equation.

cos6x=2 cos 2 3x−1 [∵cos2θ = 2cos 2 θ−1] =2 ( 4 cos 3 x−3cosx ) 2 −1 [∵cos3θ = 4cos 3 θ−3cosθ] =2(16 cos 6 x+9 cos 2 x−24 cos 4 x)−1 =32 cos 6 x+18 cos 2 x−48 cos 4 x−1

Thus, L.H.S and R.H.S of the equation are equal.