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Math Help - Proof of trig identities

  1. #1
    Junior Member
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    Unhappy Proof of trig identities

    cos(x+y) x cos(x-y) = cos squared y - sin squared x
    = cos squared x - sin squared y

    used double angle formulae except seemed to only get up to
    cossquaredx cossquared y - sin squared x sin squared y ??


    any help is GREATLY APPRECIATED

    thankss !!
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  2. #2
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    Hello, iiharthero!

    Prove: . \cos(x+y)\cos(x-y) \:=\: \cos^2\!y - \sin^2\!x \:=\:\cos^2\!x - \sin^2\!y

    Used compound-angle formulae except seemed to only get up to:

    . . \cos^2\!x\cos^2\!y - \sin^2\!x\sin^2\!y . .
    This is correct -- Good work!

    \text{We have: }\;\,\cos^2\!x\underbrace{\cos^2\!y} \quad\; - \;\quad\sin^2\!x\underbrace{\sin^2\!y}
    . . . . = \;\cos^2\!x\overbrace{(1-\sin^2\!y)} - \sin^2y\overbrace{(1 - \cos^2\!x)}

    . . . . = \;\cos^2\!x - \cos^2\!x\sin^2\!y - \sin^2\!y + \cos^2\!x\sin^2\!y

    . . . . =\;\cos^2\!x - \sin^2\!y


    Got it?

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  3. #3
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    THANKU !
    yupp that was a great helpp, realli realli appreciate it !
    thanx for takin da time to help me =DD
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by iiharthero View Post
    THANKU !
    yupp that was a great helpp =DD
    thank you.
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