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Math Help - Trig equations and proof of identity

  1. #1
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    Trig equations and proof of identity

    Can anyone help with this question?



    I have done all of the parts except the last bit, kinda went a bit over my head there :S

    There is also this question that I have gotten really stumped with, I have to try and prove the identitiy, we have only covered the first trig identity so far but we have been set this question, so I am not quite sure what I can do :S

    1+\tan^2\theta=\frac{1}{\cos^2\theta}
    Last edited by rhysiboy; April 17th 2010 at 12:29 PM.
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  2. #2
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    What did you get for part d) ?

    Look closely between the relationship in the equations in parts d) and e)

    There is no difference given x = 2\theta
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  3. #3
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    Quote Originally Posted by pickslides View Post
    What did you get for part d) ?

    Look closely between the relationship in the equations in parts d) and e)

    There is no difference given x = 2\theta
    Okay so with part d) I got solutions to = 0, pi and 2pi solutions which I got from: (x=cox^-1..{from cos^2x=1}

    So basically my solutions to part d would be the same as x = 2\theta ?
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  4. #4
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    Quote Originally Posted by rhysiboy View Post
    Can anyone help with this question?



    I have done all of the parts except the last bit, kinda went a bit over my head there :S

    There is also this question that I have gotten really stumped with, I have to try and prove the identitiy, we have only covered the first trig identity so far but we have been set this question, so I am not quite sure what I can do :S

    1+\tan^2\theta=\frac{1}{\cos^2\theta}
    1+\tan^2\theta=\frac{1}{\cos^2\theta}

    1+\frac{sin^2\theta}{\cos^2\theta}=\frac{1}{\cos^2  \theta}

    Do you see the proof now?
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  5. #5
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    Quote Originally Posted by harish21 View Post
    1+\tan^2\theta=\frac{1}{\cos^2\theta}

    1+\frac{sin^2\theta}{\cos^2\theta}=\frac{1}{\cos^2  \theta}

    Do you see the proof now?
    okay so what I did was:

    1+\frac{sin^2\theta}{\cos^2\theta}=\frac{1}{\cos^2  \theta}

    So I multiplied the {\cos^2\theta} on the left side to the right side, leaving me with 1+sin^2\theta=\frac{1*\cos^2\theta}{\cos^2\theta}

    simplified: 1+sin^2\theta=1 so effectively sin^2\theta=0... That proves the identitiy doesn't it?
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  6. #6
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    Quote Originally Posted by rhysiboy View Post

    simplified: 1+sin^2\theta=1 so effectively sin^2\theta=0... That proves the identitiy doesn't it?
    No it doesn't

    When working with a proof you must alter only one side of the equation to equate the other.

    As harish21 has already shown and only changing the LHS

    1+\tan^2\theta

    =1+\frac{\sin^2\theta}{\cos^2\theta}

    =\frac{\cos^2\theta}{\cos^2\theta}+\frac{\sin^2\th  eta}{\cos^2\theta}

    =\frac{\cos^2\theta}{\cos^2\theta}+\frac{1-\cos^2\theta}{\cos^2\theta}

    =\frac{\cos^2\theta+1-\cos^2\theta}{\cos^2\theta}

    =\frac{1}{\cos^2\theta}

    as required.
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