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Math Help - Trig university question

  1. #1
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    Unhappy Trig university question

    Hey I'm currently doing a math bridging course in university and cant seem to remember a lot of the math i learnt in school

    the question is

    prove that tan^2 theta - sin^2 theta = tan^2 theta*sin^2 theta

    sorry i dont have math equation writing program..it looks a bit weird in words.. any help would be greatly appreciated

    thanks again
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  2. #2
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    Quote Originally Posted by smplease View Post
    Hey I'm currently doing a math bridging course in university and cant seem to remember a lot of the math i learnt in school

    the question is

    prove that tan^2 theta - sin^2 theta = tan^2 theta*sin^2 theta

    sorry i dont have math equation writing program..it looks a bit weird in words.. any help would be greatly appreciated

    thanks again
    Left hand side = \tan^2 \theta - \sin^2 \theta


    = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta


    =\frac{\sin^2 \theta}{\cos^2 \theta} - \frac{\sin^2 \theta \, \cos^2 \theta}{\cos^2 \theta}


    =\frac{\sin^2 \theta - \sin^2 \theta \, \cos^2 \theta}{\cos^2 \theta}


    =\frac{\sin^2 \theta (1 - \cos^2 \theta)}{\cos^2 \theta}


    =\frac{\sin^2 \theta (\sin^2 \theta)}{\cos^2 \theta}


    =\frac{\sin^2 \theta}{\cos^2 \theta} \times \sin^2 \theta


    \tan^2 \theta \times \sin^2 \theta


    = Right Hand Side.

    Q.E.D.
    Last edited by mr fantastic; March 9th 2008 at 09:24 PM. Reason: Added a missing theta (lost a tree in the forest)
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  3. #3
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    Smile

    Quote Originally Posted by mr fantastic View Post
    Left hand side = \tan^2 \theta - \sin^2 \theta


    = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta


    =\frac{\sin^2 \theta}{\cos^2 \theta} - \frac{\sin^2 \cos^2 \theta}{\cos^2 \theta}


    =\frac{\sin^2 \theta - sin^2 \cos^2 \theta}{\cos^2 \theta}


    =\frac{\sin^2 \theta (1 - \cos^2 \theta)}{\cos^2 \theta}


    =\frac{\sin^2 \theta (\sin^2 \theta)}{\cos^2 \theta}


    =\frac{\sin^2 \theta}{\cos^2 \theta} \times \sin^2 \theta


    \tan^2 \theta \times \sin^2 \theta


    = Right Hand Side.

    Q.E.D.

    Thanks so much for the reply and the answer I understand most of it except for the 4th line down..

    where does the 1 in (1- cos) come from?

    thanks again
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  4. #4
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    Quote Originally Posted by smplease View Post
    Thanks so much for the reply and the answer I understand most of it except for the 4th line down..

    where does the 1 in (1- cos) come from?

    thanks again
    \sin^2 \theta gets taken out as a common factor:


    <br />
=\frac{{\color{red}1} \sin^2 \theta - \sin^2 \theta \, {\color{red}\cos^2 \theta}}{\cos^2 \theta}<br />


    <br />
=\frac{\sin^2 \theta ({\color{red}1 - \cos^2 \theta})}{\cos^2 \theta}<br />
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  5. #5
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by mr fantastic View Post
    \sin^2 \theta gets taken out as a common factor:


    <br />
=\frac{{\color{red}1} \sin^2 \theta - \sin^2 {\color{red}\cos^2 \theta}}{\cos^2 \theta}<br />


    <br />
=\frac{\sin^2 \theta ({\color{red}1 - \cos^2 \theta})}{\cos^2 \theta}<br />
    Maybe he was just a bit confused because the second \sin^2{\theta} didn't have a \theta in it.
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  6. #6
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    Quote Originally Posted by DivideBy0 View Post
    Maybe he was just a bit confused because the second \sin^2{\theta} didn't have a \theta in it.
    Thanks for that. I only just noticed then when you pointed it out. I hope it was obvious a theta was meant to be there.
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  7. #7
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    Thumbs up

    Quote Originally Posted by mr fantastic View Post
    Thanks for that. I only just noticed then when you pointed it out. I hope it was obvious a theta was meant to be there.
    my eyes are now open thanks so much for the help i understand it now
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