Question

How do you verify `(2cos^2x-1)^2/(cos^4x-sin^4x)=1-2sin^2x`?

Answer 1

Cosine double angle identity:
`cos(2x)=cos^2(x)-sin^2(x)`

Next, substitute rearranged `sin^2(x)+cos^2(x)=1`

Substitute `sin^2(x) = 1-cos^2(x)`:
`2cos^2(x)-1`
Sub `cos^2(x)`:
`1-2sin^2(x)`

`((cos(2x))^2)/(cos^4(x)-sin^4(x)) = cos(2x)`

The denominator `sin^4(x)-cos^4(x)` is a difference of squares:
`(sin^2(x)-cos^2(x))(sin^2(x)+cos^2(x))`
`=(cos(2x))*1`

Therefore:

`cos^2(2x)/cos(2x) = cos(2x)`

Factors cancel, leaving:
`cos(2x) = cos(2x)`