# Finding a composite inverse function.

• May 1st 2010, 10:11 PM
JennyFlowers
Finding a composite inverse function.
Hey guys! I need to find:

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

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:

http://i43.tinypic.com/s2e5bc.jpg

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!
• May 1st 2010, 10:29 PM
harish21
Quote:

Originally Posted by JennyFlowers
Hey guys! I need to find:

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

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:

http://i43.tinypic.com/s2e5bc.jpg

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

$\therefore tan(cot^{-1}(\frac{-1}{\sqrt 3})) = -\sqrt{3}$
• May 2nd 2010, 04:39 AM
JennyFlowers
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!
• May 2nd 2010, 07:11 AM
Soroban
Hello, JennyFlowers!

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

Quote:

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

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]

\csc^{-1}(\sqrt{2}) &=& \sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right) \end{array}$

• May 2nd 2010, 07:47 AM
JennyFlowers
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?
• May 2nd 2010, 08:34 AM
HallsofIvy
"Tangent" is periodic with period $\pi$. Drawing your angle in quadrant IV rather than quadrant II was just adding $\pi$ radians to it.

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