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Thread: Finding Inverse Trig Functions

  1. #1
    Member RedBarchetta's Avatar
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    Finding Inverse Trig Functions

    I know this may sound elementary...but trig has always been a frustration of mine. What are some good tricks to know or techniques on figuring these out? I mean I know all the common angles in the first quadrant, but I'm not well versed after that. I think therein lies a problem. Is it just a matter of memorizing the unit circle, restricted domains of the inverses, and the following identities?

    $\displaystyle
    \begin{gathered}
    \cos ^{ - 1} x = \pi /2 - \sin ^{ - 1} x \hfill \\
    \cot ^{ - 1} x = \pi /2 - \tan ^{ - 1} x \hfill \\
    \csc ^{ - 1} x = \pi /2 - \sec ^{ - 1} x \hfill \\
    \sin ^{ - 1} ( - x) = - \sin ^{ - 1} (x) \hfill \\
    \cos ^{ - 1} ( - x) = \pi - \cos ^{ - 1} x \hfill \\
    \end{gathered}
    $

    For example. I know this is wrong but this what I tend to do.

    $\displaystyle
    \begin{gathered}
    \cot ^{ - 1} ( - 1) \hfill \\
    \cot \theta = - 1 \hfill \\
    \frac{1}
    {{\tan \theta }} = - 1 \hfill \\
    \tan \theta = - 1 \hfill \\
    \theta = - \pi /4 \hfill \\
    \end{gathered}
    $

    Until I plug into the above formula, do I discover I'm wrong.

    $\displaystyle
    \begin{gathered}
    \cot ^{ - 1} ( - 1) = \pi /2 + \tan ^{ - 1} 1 \hfill \\
    \cot ^{ - 1} ( - 1) = \pi /2 + \pi /4 \hfill \\
    \cot ^{ - 1} ( - 1) = 3\pi /4 \hfill \\
    \end{gathered}
    $

    Is there any way to avoid these fallacies? or is it to just memorize the above?
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  2. #2
    Moo
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    Quote Originally Posted by RedBarchetta View Post
    I know this may sound elementary...but trig has always been a frustration of mine. What are some good tricks to know or techniques on figuring these out? I mean I know all the common angles in the first quadrant, but I'm not well versed after that. I think therein lies a problem. Is it just a matter of memorizing the unit circle, restricted domains of the inverses, and the following identities?

    $\displaystyle
    \begin{gathered}
    \cos ^{ - 1} x = \pi /2 - \sin ^{ - 1} x \hfill \\
    \cot ^{ - 1} x = \pi /2 - \tan ^{ - 1} x \hfill \\
    \csc ^{ - 1} x = \pi /2 - \sec ^{ - 1} x \hfill \\
    \sin ^{ - 1} ( - x) = - \sin ^{ - 1} (x) \hfill \\
    \cos ^{ - 1} ( - x) = \pi - \cos ^{ - 1} x \hfill \\
    \end{gathered}
    $

    For example. I know this is wrong but this what I tend to do.

    $\displaystyle
    \begin{gathered}
    \cot ^{ - 1} ( - 1) \hfill \\
    \cot \theta = - 1 \hfill \\
    \frac{1}
    {{\tan \theta }} = - 1 \hfill \\
    \tan \theta = - 1 \hfill \\
    \theta = - \pi /4 \hfill \\
    \end{gathered}
    $

    Until I plug into the above formula, do I discover I'm wrong.

    $\displaystyle
    \begin{gathered}
    \cot ^{ - 1} ( - 1) = \pi /2 + \tan ^{ - 1} 1 \hfill \\
    \cot ^{ - 1} ( - 1) = \pi /2 + \pi /4 \hfill \\
    \cot ^{ - 1} ( - 1) = 3\pi /4 \hfill \\
    \end{gathered}
    $

    Is there any way to avoid these fallacies? or is it to just memorize the above?
    Well, your reasoning is completely correct !

    The only problem is that you applied the wrong formula ^^

    $\displaystyle \cot^{-1}(x)=\left\{\begin{array}{ll} -\tfrac \pi 2-\tan^{-1}(x) \quad \text{if } x<0 \\ \tfrac \pi 2-\tan^{-1}(x) \quad \text{if } x \ge 0 \end{array} \right.$

    and you used $\displaystyle \tfrac \pi 2 {\color{red}+} \tan^{-1}(x)$ ?

    And don't forget x=-1, not 1.


    As for the matter of learning these formulae or not... I don't think it's something you have to know by heart. Your reasoning is perfect. You just have to be careful with the domain of the inverses. I guess it would be enough to know it (and of course, to know how to use the unit circle, but this is a mandatory condition)... See how you thought you were wrong ?
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  3. #3
    Member RedBarchetta's Avatar
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    Quote Originally Posted by Moo View Post
    and you used $\displaystyle \tfrac \pi 2 {\color{red}+} \tan^{-1}(x)$ ?
    $\displaystyle
    \tan ^{ - 1} ( - x) = - \tan ^{ - 1} (x)
    $

    Isn't that identity true? If so:

    $\displaystyle
    \begin{gathered}
    \cot ^{ - 1} ( - 1) = \pi /2 - \tan ^{ - 1} ( - 1) \hfill \\
    \cot ^{ - 1} ( - 1) = \pi /2 - ( - 1)\tan ^{ - 1} (1) \hfill \\
    \cot ^{ - 1} ( - 1) = \pi /2 + \tan ^{ - 1} (1) \hfill \\
    \end{gathered}
    $
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  4. #4
    Moo
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    Quote Originally Posted by RedBarchetta View Post
    $\displaystyle
    \tan ^{ - 1} ( - x) = - \tan ^{ - 1} (x)
    $

    Isn't that identity true? If so:

    $\displaystyle
    \begin{gathered}
    \cot ^{ - 1} ( - 1) = \pi /2 - \tan ^{ - 1} ( - 1) \hfill \\
    \cot ^{ - 1} ( - 1) = \pi /2 - ( - 1)\tan ^{ - 1} (1) \hfill \\
    \cot ^{ - 1} ( - 1) = \pi /2 + \tan ^{ - 1} (1) \hfill \\
    \end{gathered}
    $
    Okay

    The thing is that the formula is $\displaystyle {\color{red}-} \tfrac \pi 2-\tan^{-1}(x)$ for x<0 (I looked it up here http://mathworld.wolfram.com/InverseCotangent.html - (10) and (15) combined), which is the case here >.<
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  5. #5
    Member RedBarchetta's Avatar
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    Quote Originally Posted by Moo View Post
    Okay

    The thing is that the formula is $\displaystyle {\color{red}-} \tfrac \pi 2-\tan^{-1}(x)$ for x<0 (I looked it up here Inverse Cotangent -- from Wolfram MathWorld - (10) and (15) combined), which is the case here >.<
    Hmmm. Alright:

    $\displaystyle
    \begin{gathered}
    \cot ^{ - 1} ( - 1) = - \pi /2 - \tan ^{ - 1} ( - 1) \hfill \\
    \cot ^{ - 1} ( - 1) = - \pi /2 + \pi /4 \hfill \\
    \cot ^{ - 1} ( - 1) = - \pi /4 \hfill \\
    \end{gathered}
    $

    My answer says 3pi/4. So what is right?
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  6. #6
    Moo
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    Well, a confusion can come from here (continuing your working) :

    $\displaystyle \tan \theta=-1$

    If $\displaystyle \theta=-\tfrac \pi 4$ or if $\displaystyle \theta=\tfrac{3 \pi}{4}$, it works. Note that this is only an implication, not an equivalence.

    However, $\displaystyle \tan \theta=-1 \Leftrightarrow \theta=\tan^{-1}(-1)$

    and $\displaystyle \tan^{-1}(-1)=-\tfrac \pi 4$
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  7. #7
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    Quote Originally Posted by RedBarchetta View Post
    I know this may sound elementary...but trig has always been a frustration of mine. What are some good tricks to know or techniques on figuring these out? I mean I know all the common angles in the first quadrant, but I'm not well versed after that. I think therein lies a problem. Is it just a matter of memorizing the unit circle, restricted domains of the inverses, and the following identities?

    $\displaystyle
    \begin{gathered}
    \cos ^{ - 1} x = \pi /2 - \sin ^{ - 1} x \hfill \\
    \cot ^{ - 1} x = \pi /2 - \tan ^{ - 1} x \hfill \\
    \csc ^{ - 1} x = \pi /2 - \sec ^{ - 1} x \hfill \\
    \sin ^{ - 1} ( - x) = - \sin ^{ - 1} (x) \hfill \\
    \cos ^{ - 1} ( - x) = \pi - \cos ^{ - 1} x \hfill \\
    \end{gathered}
    $

    For example. I know this is wrong but this what I tend to do.

    $\displaystyle
    \begin{gathered}
    \cot ^{ - 1} ( - 1) \hfill \\
    \cot \theta = - 1 \hfill \\
    \frac{1}
    {{\tan \theta }} = - 1 \hfill \\
    \tan \theta = - 1 \hfill \\
    \theta = - \pi /4 \hfill \\
    \end{gathered}
    $

    Until I plug into the above formula, do I discover I'm wrong.

    $\displaystyle
    \begin{gathered}
    \cot ^{ - 1} ( - 1) = \pi /2 + \tan ^{ - 1} 1 \hfill \\
    \cot ^{ - 1} ( - 1) = \pi /2 + \pi /4 \hfill \\
    \cot ^{ - 1} ( - 1) = 3\pi /4 \hfill \\
    \end{gathered}
    $

    Is there any way to avoid these fallacies? or is it to just memorize the above?
    You're making things more confusing than they need to be.

    $\displaystyle \cot^{-1} (-1) = \theta \Rightarrow \cot \theta = -1$.

    So the possibilities are $\displaystyle \theta = -\frac{\pi}{4}$ or $\displaystyle \theta = \frac{3 \pi}{4}$.

    But the range of $\displaystyle \cot^{-1} (x)$ is $\displaystyle [0, \, \pi]$. See Inverse Trigonometric Functions or 7. The Inverse Trigonometric Functions

    So $\displaystyle \theta = \frac{3 \pi}{4}$ and so $\displaystyle \cot^{-1} (-1) = \frac{3 \pi}{4}$.


    However ......

    $\displaystyle \tan^{-1} (-1) = \theta \Rightarrow \tan \theta = -1$.

    So the possibilities are $\displaystyle \theta = -\frac{\pi}{4}$ or $\displaystyle \theta = \frac{3 \pi}{4}$.

    But the range of $\displaystyle \tan^{-1} (x)$ is $\displaystyle \left[-\frac{\pi}{2}, \, \frac{\pi}{2}\right]$.

    So $\displaystyle \theta = - \frac{\pi}{4}$ and so $\displaystyle \tan^{-1} (-1) = -\frac{\pi}{4}$.

    A different answer is not surprising since asking the value of $\displaystyle \cot^{-1} (-1)$ is a different question to asking the value of $\displaystyle \tan^{-1} (-1)$.

    I can't reconcile these results with the formula from mathworld at the moment.

    Edit: Note that when you add the graphs of $\displaystyle y = \cot^{-1} (x)$ and $\displaystyle y = \tan^{-1} (x)$ you get the graph $\displaystyle y = \frac{\pi}{2}$.
    Last edited by mr fantastic; Sep 9th 2008 at 04:05 AM.
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  8. #8
    Senior Member pankaj's Avatar
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    Hi,Mr.Fantastic,the range of $\displaystyle tan^{-1} x$ is $\displaystyle (-\frac{\pi}{2},\frac{\pi}{2})$ and that of $\displaystyle cot^{-1} x$ is $\displaystyle (0,\pi)$
    $\displaystyle
    cot^{-1} (\frac{1}{x}) = tan^{-1} x
    $ if x > 0

    $\displaystyle
    cot^{-1} (\frac{1}{x}) = tan^{-1} x + \pi
    $ if x < 0

    $\displaystyle
    tan^{-1} (\frac{1}{x}) = cot^{-1} x
    $ if x > 0

    $\displaystyle
    tan^{-1} (\frac{1}{x}) = cot^{-1} x - \pi
    $ if x < 0

    Also,$\displaystyle cot^{-1} {(-x)} = \pi -cot^{-1} x$
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