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Thread: inverse trigonometric function

  1. #1
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    inverse trigonometric function

    Sir / Madam,

    In the following problem:

    solve:

    Cos(tan-1x) = Sin[cot-1(3/4)]


    In the above said problem, I proceeded

    sin[(Pi/2) - tan-1x] = Sin[cot-1(3/4)]

    (Pi/2) - tan-1x = cot-1(3/4)

    Cot-1x = cot-1(3/4)

    x = 3/4

    However, they have given x = + 3/4 or - 3/4

    How to identify it has two answers?

    How to find the second answer.

    Kindly enlighten me.

    with warm regards,

    Aranga
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  2. #2
    Junior Member Cervesa's Avatar
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    Re: inverse trigonometric function

    $\sin[\cot^{-1}(3/4)] = 4/5$

    $\cos(\tan^{-1}{x}) = \dfrac{1}{\sqrt{1+x^2}}$


    $\dfrac{1}{\sqrt{1+x^2}} = \dfrac{4}{5} \implies 1+x^2 = \dfrac{25}{16} \implies x = \pm \dfrac{3}{4}$
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: inverse trigonometric function

    As the cosine functiion is even and the inverse tangent function is odd, we may state:

    $\displaystyle \cos\left(\arctan(x)\right)=\cos\left(\arctan(\pm x)\right)$
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