# Finding Inverse Trig Functions

• Sep 8th 2008, 10:21 PM
RedBarchetta
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?
• Sep 8th 2008, 10:35 PM
Moo
Quote:

Originally Posted by RedBarchetta
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 $\displaystyle \tfrac \pi 2 {\color{red}+} \tan^{-1}(x)$ ? (Smirk)

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 ?
• Sep 8th 2008, 10:42 PM
RedBarchetta
Quote:

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

$\displaystyle \tan ^{ - 1} ( - x) = - \tan ^{ - 1} (x)$

Isn't that identity true? If so:

$\displaystyle \begin{gathered} \cot ^{ - 1} ( - 1) = \pi /2 - \tan ^{ - 1} ( - 1) \hfill \\ \cot ^{ - 1} ( - 1) = \pi /2 - ( - 1)\tan ^{ - 1} (1) \hfill \\ \cot ^{ - 1} ( - 1) = \pi /2 + \tan ^{ - 1} (1) \hfill \\ \end{gathered}$
• Sep 8th 2008, 10:48 PM
Moo
Quote:

Originally Posted by RedBarchetta
$\displaystyle \tan ^{ - 1} ( - x) = - \tan ^{ - 1} (x)$

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 :p

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 >.<
• Sep 8th 2008, 10:55 PM
RedBarchetta
Quote:

Originally Posted by Moo
Okay :p

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? (Wondering)
• Sep 8th 2008, 11:04 PM
Moo
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 $\displaystyle \tan^{-1}(-1)=-\tfrac \pi 4$
• Sep 9th 2008, 12:55 AM
mr fantastic
Quote:

Originally Posted by RedBarchetta
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 $\displaystyle y = \frac{\pi}{2}$.
• Sep 14th 2008, 09:15 AM
pankaj
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,$\displaystyle cot^{-1} {(-x)} = \pi -cot^{-1} x$