Trig identities Proove that L.S. = R.S.

cos4X - sin4X = 1 - 2sin2X

cos^4(x) - sin^4(x) = 1 - 2sin^2(x)

Alright, work with the left side and factor it as a difference of squares:

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

Now remember that cos^2(x) + sin^2(x) = 1:

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

Now apply cos^2(x) = 1 - sin^2(x) (the same identity as in the last step, but isolating cos^2(x)):

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

Thus LS = RS and QED.

Done!

I'm assuming you meant:

cos^4(x) - sin^4(x) = 1 - 2*sin²(x)

The LHS simplifies to:

(cos²(x) - sin²(x))(sin²(x) + cos²(x))

We know that sin²(x) + cos²(x) = 1, so LHS becomes:

cos²(x) - sin²(x) = 1 - sin²(x) - sin²(x) = 1 - 2*sin²(x) = RHS

Proof:
LHS=
cos^4x-sin^4x=
(cos^2x)^2-(sin^2x)^2=
(cos^2x+sin^2x)(cos^2x-sin^2x)=
(1-sin^2x-sin^2x)=
1-2sin^2x=
RHS