Question

Prove that cos 6x = 32 cos'x-48 cos4x + cos²x-1​

Answer 1

Using,

cos2x=2cos2x−1

cos3x=4cos3x−3cosx

LHS:

cos6x=2cos23x−1

=2(4cos3x−3cosx)2−1

=2(16cos6x+9cos2x−24cos4x)−1

=32cos6x−48cos4x+

Answer 2

Answer:

the answer is explained in the picture