[SOLVED] Another trigonometric Identity

**F.A.S.T.** · March 1st 2010, 06:08 PM

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!

**Grandad** · March 2nd 2010, 12:27 AM

Hello F.A.S.T.

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:

$\sin^4x-\cos^4x = 2\sin^2x-1$

The proof, and the explanation of the steps, is:

$\sin^4x-\cos^4x = (\sin^2x)^2-(\cos^2x)^2$ , since $a^4 = (a^2)^2$ , where $a$ can be anything you like

$=(\sin^2x+\cos^2x)(\sin^2x-\cos^2x)$ , since $a^2-b^2 = (a+b)(a-b)$ - sometimes called 'the difference of two squares'

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

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

$=2\sin^2x-1$

Do you follow it all now?

Grandad

**F.A.S.T.** · March 2nd 2010, 04:43 AM

Ya, I got it!
The only step it did not get before was the 'the difference of two squares'.
Thank you very much!

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