Question

How do you prove `cos^4theta-sin^4theta=cos2theta`?

Answer 1

We'll use `rarrsin^2x+cos^2x=1`, `a^2-b^2=(a+b)(a-b)` and `cos^2x-sin^2x=cos2x`.

`LHS=cos^4x-sin^4x`

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

`=(cos^2x+sin^2x)*(cos^2x-sin^2x)`

`=1*cos2x=cos2x=RHS`