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Math Help - Express this as an algebraic equation..

  1. #1
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    Express this as an algebraic equation..

    Express sin(tan^{-1}\frac{1}{x} + cos^{-1}x) as an algebraic expression in x if x is negative.

    I don't understand what this question is asking me...

    sorry I'm asking so many questions. The teacher gave us a list of like 250 problems that are "possible" on the test and so theres quite a few I get stuck on...
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by circuscircus View Post
    Express sin(tan^{-1}\frac{1}{x} + cos^{-1}x) as an algebraic expression in x if x is negative.

    I don't understand what this question is asking me...

    sorry I'm asking so many questions. The teacher gave us a list of like 250 problems that are "possible" on the test and so theres quite a few I get stuck on...
    Replace x with -x:
    sin \left [ tan^{-1} \left ( \frac{1}{-x} \right ) + cos^{-1}(-x) \right ]

    Now, the tangent function is an odd function, so tan(-x) = -tan(x) \implies tan^{-1}(-x) = -tan^{-1}(x)

    and cosine is an even function, so
    cos(-x) = cos(x) \implies cos^{-1}(-x) = cos^{-1}(x),

    thus
    sin \left [ tan^{-1} \left ( \frac{1}{-x} \right ) + cos^{-1}(-x) \right ] = sin \left [- tan^{-1} \left ( \frac{1}{x} \right ) + cos^{-1}(x) \right ]

    We may possibly take this one step further:
    Sine is an odd function so sin(-x) = -sin(x), thus
    sin \left [- tan^{-1} \left ( \frac{1}{x} \right ) + cos^{-1}(x) \right ] = -sin \left [ tan^{-1} \left ( \frac{1}{x} \right ) - cos^{-1}(x) \right ]

    -Dan
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  3. #3
    MHF Contributor red_dog's Avatar
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    Let \alpha=\arctan x\Rightarrow\tan\alpha=x
    For x<0\Rightarrow\alpha\in\left(-\frac{\pi}{2},0\right) and \sin\alpha<0, \ \cos\alpha>0.

    Let \beta=\arccos x\Rightarrow\cos\beta=x
    For x<0\Rightarrow\beta\in\left(\frac{\pi}{2},\pi\righ  t) and \sin\beta>0.

    We have \displaystyle\sin\alpha=\frac{\tan\alpha}{\sqrt{1+  \tan^2\alpha}}=\frac{x}{\sqrt{1+x^2}}, \cos\alpha=\frac{1}{\sqrt{1+\tan^\alpha}}=\frac{1}  {\sqrt{1+x^2}}
    \sin\beta=\sqrt{1-\cos^2\beta}=\sqrt{1-x^2}, \ \cos\beta=x.

    Now, we have to calculate
    \sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\c  os\alpha=
    \displaystyle=\frac{x}{\sqrt{1+x^2}}\cdot x+\sqrt{1-x^2}\cdot\frac{1}{\sqrt{1+x^2}}=\frac{x^2+\sqrt{1-x^2}}{\sqrt{1+x^2}}
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