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Thread: [SOLVED] Another trigonometric Identity

  1. #1
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    [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!
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    Last edited by Grandad; Mar 2nd 2010 at 12:28 AM. Reason: Removed unnecessary blank lines
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  2. #2
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    Hello F.A.S.T.
    Quote Originally Posted by F.A.S.T. View Post
    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?

    Grandad
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  3. #3
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    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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