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Math Help - Finding a composite inverse function.

  1. #1
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    Finding a composite inverse function.

    Hey guys! I need to find:

    tan(cot^{-1}(\frac{-1}{\sqrt{3}}))

    Here's how I'm going about finding the answer:

    I read this as, "Tangent of the angle whose cotangent is negative 1 over the square root of 3."

    As I understand it, I need to draw the triangle for the inverse cotangent in quadrant 4 since it is negative. Here's my drawing:



    Now, using this drawing, I would find the tangent (and the answer to the expression) to be: -\sqrt{3}

    Did I do this correctly? I'm having a hard time with the concept of inverse functions in general, so any advice or somewhere I could read to better understand the concept would be helpful. I've read my trig book and it's not helping me very much.

    I'm doubting that it is correct because when I evaluate the expression on my TI-84, the answer doesn't match.

    Thanks for any help!
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by JennyFlowers View Post
    Hey guys! I need to find:

    tan(cot^{-1}(\frac{-1}{\sqrt{3}}))

    Here's how I'm going about finding the answer:

    I read this as, "Tangent of the angle whose cotangent is negative 1 over the square root of 3."

    As I understand it, I need to draw the triangle for the inverse cotangent in quadrant 4 since it is negative. Here's my drawing:





    Now, using this drawing, I would find the tangent (and the answer to the expression) to be: -\sqrt{3}

    Did I do this correctly? I'm having a hard time with the concept of inverse functions in general, so any advice or somewhere I could read to better understand the concept would be helpful. I've read my trig book and it's not helping me very much.

    I'm doubting that it is correct because when I evaluate the expression on my TI-84, the answer doesn't match.

    Thanks for any help!
    Actually,
    cot^{-1}(\frac{-1}{\sqrt 3}) returns you an angle

    so, cot^{-1}(\frac{-1}{\sqrt 3}) = \frac{-\pi}{3}

    and tan(\frac{-\pi}{3})= -tan(\frac{\pi}{3})=  -\sqrt{3}

    \therefore tan(cot^{-1}(\frac{-1}{\sqrt 3})) = -\sqrt{3}
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  3. #3
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    Out of curiosity, what difference would this make?

    tan^{-1}(cot(\frac{-1}{\sqrt{3}}))

    Does switching which function is the inverse change the outcome?

    Thanks for the help!
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  4. #4
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    Hello, JennyFlowers!

    With a little thought, this is an "eyeball" problem.
    (You get the answer by looking at the problem.)


    Find: . \tan\left[\cot^{-1}\left(-\tfrac{1}{\sqrt{3}}\right)\right]

    Here's how I'm going about finding the answer:

    I read this as: "Tangent of the angle whose cotangent is -\tfrac{1}{\sqrt{3}} " . Good!

    \text{Let }\theta \,=\,\cot^{-1}\left(-\tfrac{1}{\sqrt{3}}\right) \quad \hdots \quad \theta\text{ is an angle whose cotangent is }-\tfrac{1}{\sqrt{3}}

    Then: . \cot\theta \:=\:-\frac{1}{\sqrt{3}} \:=\:\frac{adj}{opp}

    The right triangle looks like this:
    Code:
                      *
                    * |
                  *   |  _
                *     | √3
              *       |
            * θ       |
          * - - - - - *
               -1

    And we see that: . \tan\theta \:=\:\frac{\sqrt{3}}{\text{-}1} \:=\:\text{-}\sqrt{3}

    . . Hence: . \theta \:=\:\tan^{-1}\left(\text{-}\sqrt{3}\right)



    Do you see what we've done?

    We've shown that: . \cot^{-1}\left(\text{-}\tfrac{1}{\sqrt{3}}\right) \;=\;\tan^{-1}\left(\text{-}\sqrt{3}\right)

    Give it some thought . . . The reasoning should become clear. .**



    So we can replace \cot^{-1}\left(\text{-}\tfrac{1}{\sqrt{3}}\right) with \tan^{-1}\left(\text{-}\sqrt{3}\right)

    And the original problem becomes: . \tan\left(\tan^{-1}(\text{-}\sqrt{3})\right)

    . . which equals (of course): . -\sqrt{3}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    **

    The same reasoning that allows us to say:

    . . \begin{array}{ccc}\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right) &=& \cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right) \\ \\[-2mm]<br /> <br />
\csc^{-1}(\sqrt{2}) &=& \sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right) \end{array}

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  5. #5
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    Thank you, Soroban, I think I understand now. But how did you determine that the angle should be drawn in quadrant II? Was it just luck that I got the same answer by drawing in quadrant IV?
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  6. #6
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    "Tangent" is periodic with period \pi. Drawing your angle in quadrant IV rather than quadrant II was just adding [itex]\pi[/itex] radians to it.

    In any case, since tan(x)= 1/cot(x), tan(cot^{-1}(-\frac{1}{\sqrt{3}})= 1/cot(cot^{-1}(\frac{1}{\sqrt{3}}))= -1/\frac{1}{\sqrt{3}}= -\sqrt{3}.
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