# Inverse Trigonometric Functions

• Dec 5th 2007, 08:48 PM
Truthbetold
Inverse Trigonometric Functions
y= arcsin x
Find the right end behavior model, the left hand behavior model, and any horizontal tangents for the function if they exist.

The tangent I can find easily with a calculator, but the behaviors I don't know.

I also need to do it for arccos x, though I think I should be able to get it after knowing arcsin x.
• Dec 5th 2007, 09:55 PM
earboth
Quote:

Originally Posted by Truthbetold
y= arcsin x
Find the right end behavior model, the left hand behavior model, and any horizontal tangents for the function if they exist.

The tangent I can find easily with a calculator, but the behaviors I don't know.

I also need to do it for arccos x, though I think I should be able to get it after knowing arcsin x.

Hello,

only functions with unchanging monotony (I know that this expression is not appropriate - I mean the first derivative is allways positive or allways negative) have inverse functions. The sin-function has to be bounded to be monotonously increasing (or decreasing). Usually you take the interval $\left [-\frac{\pi}{2}, \frac{\pi}{2}\right ]$.
Thus you have:

$\begin{array}{l}x\mapsto \sin(x), \ \left [-\frac{\pi}{2}, \frac{\pi}{2}\right ] \mapsto [-1, 1] \\ \ \\x\mapsto \arcsin(x), \ [-1, 1] \mapsto \left [-\frac{\pi}{2}, \frac{\pi}{2}\right ] \end{array}$

Therefore examine the arcsin-function if x appraches -1 from the right and when x approaches +1 from the left.
• Dec 6th 2007, 05:42 PM
Truthbetold
So the right is -pi/2 - because the limit as x--> -1 from the right is -pi/2
and the left is pi/2- because the limit as x--> 1 from the left is pi/2.
The answer is pi/2 for both.

What I did I do wrong?:confused: