Trig Help: Prove the identity

I'm having a tough time with this one. Please show work. Irrelevant and obnoxious answers will be reported.

Prove the identity: cosx^4 - sinx^4 = 2cosx^2-1

cosx^4 - sinx^4

= ( cosx^2 + sinx^2 ) * ( cosx^2 - sinx^2 ).............using a^2 - b^2 = (a+b)(a-b)

= ( cosx^2 - sinx^2 ).....................since ( cosx^2 + sinx^2 ) = 1

= ( cosx^2 - [1-cosx^2] )

= 2cosx^2-1

hence proved

"Irrelevant and obnoxious answers will be reported." Good for you!

(cosx)^4 - (sinx)^4 = 2(cosx)^2 - 1

LHS is the difference of two squares:
(cos²x + sin²x)(cos²x - sin²x) = 2cos²x - 1

Apply the Pythagorean Identity: (cos²x + sin²x) = 1
(1)(cos²x - sin²x) = 2cos²x -1

Apply the Pythagorean Identity: sin²x = 1-cos²x
cos²x - (1-cos²x) = 2cos²x - 1

Collect like terms in the LHS:
cos²x - 1 + cos²x = 2cos²x - 1
2cos²x - 1 = 2cos²x - 1
QED

cosx^4 - sinx^4 = 2cosx^2-1
let RHS = cosx^4 - sinx^4;
let LHS = 2cosx^2 - 1
=> cosx^4 - sinx^4
= (cosx^2 - sinx^2)(cosx^2 + sinx^2)
= cosx^2 - sinx^2
= cosx^2 - (1 - cosx^2)
= 2cosx^2-1
= LHS
=> RHS = LHS
As required to show.

By the way, I've never actually proved this before. That was way easier than I thought it would be.

cosx^4 - sinx^4
=([cos^2(x)+sin^2(x)][cos^2(x)-sin^2(x)…
=1*cos^2(x)-[1-cos^2(x)]
=cos^2(x)-1+cos^2(x)
=2cos^2(x)-1 prove//


note 2cos^2(x)-1 = cos(2x)