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

Open in App

Solution

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.

Suggest Corrections

Similar questions

cos 6x = 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1

cos 6 x = 32 {cos}^{6} x - 48 {cos}^{4} x + 18 {cos}^{2} x - 1

cos 6 x = 32 {cos}^{6} x - 48 {cos}^{4} x + 18 {cos}^{2} x - 1

c o s 6 x = 32 c o s^{6} x - 48 c o s^{4} x + 18 c o s^{2} x - 1

Join BYJU'S Learning Program