Prove the identity (sinx)^2 - (cosx)^2 = (sinx)^4 - (cosx)^4

A simpler way to see what is going on in this equation is
to redefine the essential terms. Specifically, let

(sinx)^2
= r

(cosx)^2 = s

The equation
can then be written as

r - s = r^2 -
s^2

We can then use the method of factorizing the
difference of two squares, namely

r^2 - s^2 = (r - s)(r +
s)

Note that the 'cross terms' rs and -rs cancel each other
out.

Therefore we can now write the equation
as

r - s = (r - s)(r + s)

From
the (Pythagorean, or, unit circle) trigonometric identity  (sinx)^2 + (cosx)^2 = 1  we
have that

r + s = 1

Our
equation then can be written as

r - s = (r - s) x
1

that is

r - s = r -
s

As this holds as true, we see that the original equation
does indeed hold true.

NB If you plot (sinx)^2 + (cosx)^2
on a graph the identity (sinx)^2 + (cosx)^2 = 1 is apparent as the curves are perfectly
symmetric in the line y = 1/2

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