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!

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- Mar 1st 2010, 06:08 PMF.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 AMGrandad
Hello F.A.S.T.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 likeDo you follow it all now?$\displaystyle =(\sin^2x+\cos^2x)(\sin^2x-\cos^2x)$, 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 $\displaystyle \cos^2x = 1-\sin^2x$

$\displaystyle =2\sin^2x-1$

Grandad - Mar 2nd 2010, 04:43 AMF.A.S.T.
Ya, I got it!

The only step it did not get before was the 'the difference of two squares'.

Thank you very much!