Question

Prove the identity `cos4θ - 4cos2θ≡ 8sin^4θ - 3` could really use some help on this?

`cos4θ - 4cos2θ≡ 8sin^4θ - 3`

Answer 1

`LHS=cos4θ - 4cos2θ`

`=2cos^2 2θ-1 - 4cos2θ`

`=2(1-2sin^2θ)^2-1 - 4cos2θ`

`=2(1-4sin^2theta+4sin^4theta)-1- 4(1-2sin^2θ)`

`=2-8sin^2theta+8sin^4theta-1- 4+8sin^2θ`

`= 8sin^4θ - 3=RHS`