Question

cos 4x = 1 -8 cos^2 x + cos^4 x ​

Answer 1

Let us consider the LHS

cos 4x

As we know, cos 2x = 2 cos²x-1

Therefore,

cos 4x = 2 cos² 2x-1

= 2(2 cos² 2x - 1)² - 1

= 2[(2 cos² 2x)² + 12 -2 × 2 cos²x] - 1 X

= 2(4 cos4 2x + 1 - 4 cos²x) - 1

= 8 cos4 2x + 2-8 cos²x - 1

= 8 cos4 2x + 1 - 8 cos²x

= RHS