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Math Help - 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?

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

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

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

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

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

    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?

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

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

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

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

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

    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 ^^

    \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 \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 \tfrac \pi 2 {\color{red}+} \tan^{-1}(x) ?
    <br />
\tan ^{ - 1} ( - x) =  - \tan ^{ - 1} (x)<br />

    Isn't that identity true? If so:

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

    Isn't that identity true? If so:

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

    The thing is that the formula is {\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 {\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:

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

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

    \tan \theta=-1

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

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

    and \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?

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

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

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

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

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

    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.

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

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

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

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


    However ......

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

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

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

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

    A different answer is not surprising since asking the value of \cot^{-1} (-1) is a different question to asking the value of \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 y = \cot^{-1} (x) and y = \tan^{-1} (x) you get the graph y = \frac{\pi}{2}.
    Last edited by mr fantastic; September 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 tan^{-1} x is (-\frac{\pi}{2},\frac{\pi}{2}) and that of cot^{-1} x is (0,\pi)
     <br />
cot^{-1} (\frac{1}{x}) = tan^{-1} x<br />
if x > 0

     <br />
cot^{-1} (\frac{1}{x}) = tan^{-1} x + \pi<br />
if x < 0

     <br />
tan^{-1} (\frac{1}{x}) = cot^{-1} x<br />
if x > 0

     <br />
tan^{-1} (\frac{1}{x}) = cot^{-1} x - \pi<br />
if x < 0

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