# [SOLVED] Another trigonometric Identity

• Mar 1st 2010, 06:08 PM
F.A.S.T.
[SOLVED] Another trigonometric Identity
I figured out the problem before, but now I have no clue what I did. Can you please help me figure this out!

The It is in the word document!
• Mar 2nd 2010, 12:27 AM
Hello F.A.S.T.
Quote:

Originally Posted by F.A.S.T.
I figured out the problem before, but now I have no clue what I did. Can you please help me figure this out!

The It is in the word document!

The question is this: Prove the identity:
$\displaystyle \sin^4x-\cos^4x = 2\sin^2x-1$
The proof, and the explanation of the steps, is:
$\displaystyle \sin^4x-\cos^4x = (\sin^2x)^2-(\cos^2x)^2$, since $\displaystyle a^4 = (a^2)^2$, where $\displaystyle a$ can be anything you like
$\displaystyle =(\sin^2x+\cos^2x)(\sin^2x-\cos^2x)$, since $\displaystyle a^2-b^2 = (a+b)(a-b)$ - sometimes called 'the difference of two squares'

$\displaystyle = 1(\sin^2x-\cos^2x)$, since $\displaystyle \sin^2x+\cos^2x=1$, for any value of $\displaystyle x$

$\displaystyle =\sin^2x - (1-\sin^2x)$, using $\displaystyle \sin^2x+\cos^2x=1$ again, but written as $\displaystyle \cos^2x = 1-\sin^2x$

$\displaystyle =2\sin^2x-1$
Do you follow it all now?