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Math Help - Prove the following equasion:

  1. #1
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    Prove the following equasion:

    Hey guys,
    I have a question to prove the following equation:
    Sin^2aCos^2B - Sin^2Bcos^2a = Cos^2B - Cos^2a<br />
    What I mean by 'proving' is that you have to take one side of the equation and reach the other side...
    Usually its easier to take the left side.
    B = Beta a = Alpha
    Cheers,
    Punkbuster
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  2. #2
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    Quote Originally Posted by Punkbuster View Post
    Hey guys,
    I have a question to prove the following equation:
    Sin^2aCos^2B - Sin^2Bcos^2a = Cos^2B - Cos^2a<br />
    What I mean by 'proving' is that you have to take one side of the equation and reach the other side...
    Usually its easier to take the left side.
    B = Beta a = Alpha
    Cheers,
    Punkbuster
    Dirty algebra trick + Algebra + trigonometric Pythagoras theorem. Read carefully the following and justify each step:

    \displaystyle{\sin^2\alpha\cos^2\beta-\sin^2\beta\cos^2\alpha=\sin^2\alpha\cos^2\beta-\sin^2\alpha\cos^2\alpha+\sin^2\alpha\cos^2\alpha-\sin^2\beta\cos^2\alpha=}

    =\displaystyle{\sin^2\alpha(\cos^2\beta-\cos^2\alpha)+\cos^2\alpha(\sin^2\alpha-\sin^2\beta)=\sin^2\alpha(cos^2\beta-\cos^2\alpha)+\cos^2\alpha(1-\cos^2\alpha-(1-\cos^2\beta))=}

    =\displaystyle{\sin^2\alpha(\cos^2\beta-\cos^2\alpha)+\cos^2\alpha(\cos^2\beta-\cos^2\alpha)}=....end the proof.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Dirty algebra trick + Algebra + trigonometric Pythagoras theorem. Read carefully the following and justify each step:

    \displaystyle{\sin^2\alpha\cos^2\beta-\sin^2\beta\cos^2\alpha=\sin^2\alpha\cos^2\beta-\sin^2\alpha\cos^2\alpha+\sin^2\alpha\cos^2\alpha-\sin^2\beta\cos^2\alpha=}

    =\displaystyle{\sin^2\alpha(\cos^2\beta-\cos^2\alpha)+\cos^2\alpha(\sin^2\alpha-\sin^2\beta)=\sin^2\alpha(cos^2\beta-\cos^2\alpha)+\cos^2\alpha(1-\cos^2\alpha-(1-\cos^2\beta))=}

    =\displaystyle{\sin^2\alpha(\cos^2\beta-\cos^2\alpha)+\cos^2\alpha(\cos^2\beta-\cos^2\alpha)}=....end the proof.

    Tonio
    Damn!! Can you believe I might need to do something like that in an exam??
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  4. #4
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    Hello, Punkbuster!

    Use the identity: . \sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \sin^2\!\theta \:=\:1-\cos^2\!\theta


    \text{Prove: }\:\sin^2\!A\cos^2\!B - \sin^2\!B\cos^2\!A \:=\:\cos^2\!B - \cos^2\!A

    \sin^2\!A\cos^2\!B - \sin^2\!B\cos^2\!A

    . . =\;(1-\cos^2\!A)\cos^2\!B - (1-\cos^2\!B)\cos^2\!A

    . . =\;\cos^2\!B - \cos^2\!A\cos^2\!B - \cos^2\!A + \cos^2\!A\cos^2\!B

    . . =\;\cos^2\!B - \cos^2\!A

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, Punkbuster!

    Use the identity: . \sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \sin^2\!\theta \:=\:1-\cos^2\!\theta



    \sin^2\!A\cos^2\!B - \sin^2\!B\cos^2\!A

    . . =\;(1-\cos^2\!A)\cos^2\!B - (1-\cos^2\!B)\cos^2\!A

    . . =\;\cos^2\!B - \cos^2\!A\cos^2\!B - \cos^2\!A + \cos^2\!A\cos^2\!B

    . . =\;\cos^2\!B - \cos^2\!A

    I was about to come here and repost what you said...
    But then I found that you raced me
    Thanks
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