Question

Question #84027

Answer 1

`32x^2-1`

Explanation:

Let `theta=cos^("-"1)(4x)`
`cos(2cos^("-"1)(4x)=cos(2theta)`

Use the double-angle formula `cos(2alpha)=2cos^2alpha-1`
`cos(2theta)=2cos^2theta-1`

`theta` is some angle the cosine of which is `4x`, so `cos^2theta=(4x)^2`
`2cos^2theta-1=2(4x)^2-1`

`2(4x)^2-1=2(16x^2)-1`
`2(16x^2)-1=32x^2-1`