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

Note that `cos^4x=(cos^2x)^2`

`L.H.S.`
`=cos^4x-sin^4x`

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

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

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

`=(1)(1-2sin^2x)`

`=1-2sin^2x`
`=R.H.S.`

To prove that `cos^4x-sin^4x=1-2sin^2x`, we'll need the Pythagorean identity and a variation on the Pythagorean identity:

`color(white)=>cos^2x+sin^2x=1`

`=>cos^2x=1-sin^2x`

I'll start with the left-hand side and manipulate it until it looks like the right-hand side using these two identities:

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

`color(white)(LHS)=(cos^2x)^2-(sin^2x)^2`

`color(white)(LHS)=(color(red)(cos^2x+sin^2x))(cos^2x-sin^2x)`

`color(white)(LHS)=color(red)1*(cos^2x-sin^2x)`

`color(white)(LHS)=color(blue)(cos^2x)-sin^2x`

`color(white)(LHS)=color(blue)(1-sin^2x)-sin^2x`

`color(white)(LHS)=1-2sin^2x`

`color(white)(LHS)=RHS`

That's the proof. Hope this helped!

To verify: `cos^4x - sin^4x = 1-2sin^2x`

Let `theta = sin^2x -> cos^2x = 1-theta`

`:. LHS = (1-theta)^2 - theta^2`

`= 1-2theta+theta^2-theta^2`

`= 1-2theta`

Undo substitution:

`LHS = 1-2sin^2x = RHS`