# Prove the identity (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),`...

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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.

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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

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