Question

Prove - cos6x = 32cos^6x - 48cos^4x + 18cos^2x-1 ​

Answer 1

Answer:

Using,

cos2x=2cos

2

x−1

cos3x=4cos

3

x−3cosx

LHS:

cos6x=2cos

2

3x−1

=2(4cos

3

x−3cosx)

2

−1

=2(16cos

6

x+9cos

2

x−24cos

4

x)−1

=32cos

6

x−48cos

4

x+18cos

2

x−1 = RHS

Answer 2

Answer:

2 cos^2(3x) -1

Explanation:

For *LHS*

cos2(3x) = *2cos^2x -1*

For *RHS*

TAKE 2 COMMON

> 2(16cos^6x - 24cos^4x + 9cos^2x) - 1

APPLY THE PROPERTY OF

(A-B)^2 = (A^2 - 2AB + B^2)

> 2(4cos^3x - 3cos^x)^2 - 1

> 2(cos3x)^2 - 1

> *2cos^2x -1*

LHS = RHS

*HENCE* *PROVED*