prove the following identity

cos^4X - sin^4X = 1-2sin^2X

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

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

Factorize the left hand side using:

A²-B²=(A+B)(A-B)
or
A4-B4 = (A²+B²)(A²-B²)

To prove the identity cos^4X - sin^4X = 1 - 2sin^2X, we will work with the left-hand side (LHS) and manipulate it until it matches the right-hand side (RHS).

Starting with the LHS:
cos^4X - sin^4X

We can rewrite cos^4X as (cos^2X)^2 and sin^4X as (sin^2X)^2:
(cos^2X)^2 - (sin^2X)^2

Now, let's apply the identity a^2 - b^2 = (a + b)(a - b):
[(cos^2X) + (sin^2X)][(cos^2X) - (sin^2X)]

We know that cos^2X + sin^2X = 1 from the Pythagorean identity:
[1][(cos^2X) - (sin^2X)]

Finally, we can simplify the expression:
cos^2X - sin^2X

Now, we can apply another trigonometric identity, cos^2X - sin^2X = cos(2X):
cos(2X)

Therefore, we have proven that the left-hand side (LHS) of the given identity simplifies to cos(2X), which matches the right-hand side (RHS) of the identity, 1 - 2sin^2X.