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Math Help - trigonometric identities problem

  1. #1
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    trigonometric identities problem

    I'm having problems in proving that cos(θ+φ) cos(θ-φ) = cos²θ-sin²θ using the trigonometric identities. i'm having cos²θ-sin²φ as the answer. help!
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  2. #2
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    cos(\theta + \phi) cos(\theta - \phi) = (cos\theta cos\phi - sin\theta sin \phi)(cos\theta cos\phi + sin\theta sin \phi)

    =cos^2\theta cos^2\phi + cos\theta sin\theta cos\phi sin\phi - cos\theta sin\theta cos\phi sin\phi - sin^2 \theta sin^2 \phi

    =cos^2\theta cos^2\phi - sin^2 \theta sin^2 \phi

    =cos^2\theta (1- sin^2\phi) - sin^2 \theta (1-cos^2 \phi)

    =cos^2\theta - cos^2\theta sin^2\phi - sin^2 \theta+ sin^2\theta cos^2 \phi

    =cos^2\theta - sin^2 \theta+ sin^2\theta cos^2 \phi - cos^2\theta sin^2\phi

    Unless sin^2\theta cos^2 \phi - cos^2\theta sin^2\phi = 0 (which I don't think is true), I think there was a mistake in the question you got.
    Last edited by Unknown008; August 15th 2010 at 06:15 AM. Reason: Ok, corrected LaTeX.
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  3. #3
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    Quote Originally Posted by khms View Post
    I'm having problems in proving that cos(θ+φ) cos(θ-φ) = cos²θ-sin²θ using the trigonometric identities. i'm having cos²θ-sin²φ as the answer. help!
    Cos(A+B)=CosACosB-SinASinB

    Cos(A-B)=Cos[A+(-B)]=CosACos(-B)-SinASin(-B)

    Cos(-B)=CosB,\ Sin(-B)=-SinB

    Cos(A-B)=CosACosB+SinASinB


    [Cos(A+B)]Cos(A-B)=[CosACosB-SinASinB][CosACosB+SinASinB]

    =Cos^2ACos^2B-Sin^2ASin^2B

    =Cos^2A\left(1-Sin^2B\right)-Sin^2ASin^2B=Cos^2A-Sin^2B\left(Cos^2A+Sin^2A\right)=Cos^2A-Sin^2B

    which is what you got!
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  4. #4
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    i know. but what i am looking for is the proof that LHS and RHS are equal. the answers i'm getting are incorrect. i need cos²θ-sin²θ as the answer - not cos²θ-sin²φ.
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  5. #5
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    Quote Originally Posted by khms View Post
    i know. but what i am looking for is the proof that LHS and RHS are equal. the answers i'm getting are incorrect. i need cos²θ-sin²θ as the answer - not cos²θ-sin²φ.
    You will get that answer if \theta=\varphi

    since Cos(2A)=Cos^2A-Sin^2A
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    You will get that answer if \theta=\varphi

    since Cos(2A)=Cos^2A-Sin^2A
    so what you're saying is it is impossible to get cos²θ-sin²θ as the answer for that equation?
    Last edited by khms; August 15th 2010 at 07:18 AM. Reason: typo
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  7. #7
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    No, what they are saying is that \cos^2{\theta} - \sin^2{\theta} is only possible if \theta = \varphi...
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  8. #8
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    There is a typo in the question !







    ... from previous poster. Continue as follows:

    = (1 - sin²A).(1-sin²B) - sin²A.sin²B (sorry can't use greek letters !)

    = 1 -sin²A - sin²B + sin²A.sin²B - sin²A.sin²B

    = 1 -sin²A - sin²B

    = sin²A + cos²A - sin²A - sin²B

    = cos²A - sin²B

    The question as written asked for the RHS to be cos²A - sin²A which is, of course, incorrect !

    Good luck,

    M
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