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**

Hello!

First, consider the right part:

`sin^4(x) - cos^4(x)`

and express it as the difference of squares:

`[sin^2(x)]^2 - [cos^2(x)]^2.`

Now use the formula a^2 - b^2 = (a-b)*(a+b) and obtain

`[sin^2(x) - cos^2(x)] *[sin^2(x) + cos^2(x)].`

But the second factor is always 1, and we get

`sin^2(x) - cos^2(x),` which is the left part of the identity. QED.

For the beginning, we'll manipulate the right side, only.

[(sin x)^2]^2 - [(cos x)^2]^2 = [(sin x)^2 - (cos x)^2][(sin x)^2 + (cos x)^2]

But, (sin x)^2 + (cos x)^2 = 1 (Pythagorean identity)

[(sin x)^2]^2 - [(cos x)^2]^2 = [(sin x)^2 - (cos x)]^2

We notice that we've get the difference of squares from the left side:

**[(sin x)^2 - (cos x)]^2 = [(sin x)^2 - (cos x)]^2**

consider the identity, `x^2 -y^2 = (x-y)(x+y)`.

using the same on the right hand side of the equation, we get:

`(sinx)^4 - (cosx)^4 = [(sinx)^2 - (cosx)^2)].[(sinx)^2 + (cosx)^2)]`

Using the identity, `(sinx)^2 + (cosx)^2) = 1`, this equation simplifies to:

`(sinx)^4 - (cosx)^4 = [(sinx)^2 - (cosx)^2)].[(sinx)^2 + (cosx)^2)] = (sinx)^2 - (cosx)^2)`

which is the same as the left hand side of the equation. and hence proved.

In general, for this type of mathematical problems, its always ideal to start by one side of the equation (either LHS or RHS, i.e. left hand side or right hand side) and simplifying it. Use of identities, if applicable, helps in simplification process. and finally we get the two sides as equal to each other, i.e., the requisite answer.

Good luck.