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Math Help - Solving trigonometric equations involving arctan?

  1. #1
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    Solving trigonometric equations involving arctan?

    I'm having difficulty solving the following equation which helps locate the angles in young's double slit experiment. Currently I'm using a graphical method.


    The equation is:

    C^2 - b^2 - a^2 = \frac{a}{\tan^2(\theta)} + \frac{2aC}{\sin(\theta)} - \frac{2ba}{\tan(\theta)} - \frac{a^2}{\sin^2(\theta)}

    a, b and C are constants.



    I tried expressing {\tan(\theta)} in terms of {\sin(\theta)}.

    However,  {\tan(\theta)} = \frac{\sin(\theta)}{\sqrt{\1-sin^2(\theta)}}

    And I'm not sure how to deal with the , given that I would like to make θ the subject, where 0 < θ < \pi.
    Last edited by niggawut; October 1st 2011 at 05:40 AM.
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  2. #2
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    Re: Solving trigonometric equations involving arctan?

    how do you express arctan in terms of basic trigonometric functions?
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Solving trigonometric equations involving arctan?

    You can use the fact that \sin\left[\arctan(x)\right]=\frac{x}{\sqrt{1+x^2}}
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    Re: Solving trigonometric equations involving arctan?

    thanks.
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    Re: Solving trigonometric equations involving arctan?

    Quote Originally Posted by Siron View Post
    You can use the fact that \sin\left[\arctan(x)\right]=\frac{x}{\sqrt{1+x^2}}
    So x is simply any angle?

    Also, how do you input math notation like that?
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Solving trigonometric equations involving arctan?

    x isn't an angle, it's a real number, for example \arctan(1)=\frac{\pi}{4}+k\pi (Have you ever worked with cyclometric functions before?)
    In this case x=\frac{a}{\left|b-\left(\frac{a}{\tan(\theta)}\right)\right|} so \sin[\arctan(x)]=...?
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  7. #7
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    Re: Solving trigonometric equations involving arctan?

    Ok so i used the theorem and ended up with:

    C^2 - b^2 - a^2 = \frac{a}{\tan^2(\theta)} + \frac{2aC}{\sin(\theta)} - \frac{2ba}{\tan(\theta)} - \frac{a^2}{\sin^2(\theta)}

    Yeah I learned how to use the notation.

    I tried expressing {\tan(\theta)} in terms of {\sin(\theta)}.

    However,  {\tan(\theta)} = \frac{\sin(\theta)}{\sqrt{\1-sin^2(\theta)}}

    And I'm not sure how to deal with the , given that I would like to make θ the subject, where 0 < θ < \pi.
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  8. #8
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    Re: Solving trigonometric equations involving arctan?

    I tried resolving it into a quadratic equation in terms of sin(\theta) , but it always gives me answers greater than 1.
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  9. #9
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    Re: Solving trigonometric equations involving arctan?

    Quote Originally Posted by niggawut View Post
    I tried resolving it into a quadratic equation in terms of sin(\theta) , but it always gives me answers greater than 1.
    I squared both sides in order to form the quadratic equation but it doesnt seem to be working.
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  10. #10
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    Re: Solving trigonometric equations involving arctan?

    perhaps my co-effs are wrong.
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