# Trigonometric inverses and their graphs?

• Jan 20th 2011, 05:09 PM
homeylova223
Trigonometric inverses and their graphs?
Write the equation for the inverse of each function. Then graph the function and it inverse.
This is the original function y=tan x/2
This is what I did
y= tan x/2
x=tan y/2
2x=tan y
y= 2 arc-tan x
But I am confused on how to graph these functions as they have pi as the asymptote while on other tan function it is zero. Can anyone show me how to graph this function as I am confused?(Crying)
• Jan 20th 2011, 05:41 PM
skeeter
Quote:

Originally Posted by homeylova223
Write the equation for the inverse of each function. Then graph the function and it inverse.
This is the original function y=tan x/2
This is what I did
y= tan x/2
x=tan y/2
2x=tan y ... no

$y = \tan\left(\frac{x}{2}\right)$

$\frac{x}{2} = \arctan{y}$

$x = 2\arctan{y}$

$y = 2\arctan{x}$
• Jan 20th 2011, 05:46 PM
homeylova223
The thing is I am supposed the graph using radians and I am confused on how I would use my unit circle as the graph on the back of my book show an asymptote at -pi and pi on this graph.
• Jan 20th 2011, 05:55 PM
skeeter
Quote:

Originally Posted by homeylova223
The thing is I am supposed the graph using radians and I am confused on how I would use my unit circle as the graph on the back of my book show an asymptote at -pi and pi on this graph.

the graphs provided are in radians ... note what the black graph ( $y = \tan\left(\frac{x}{2}\right)$ ) does as it gets close to the value $x = 3$ and $x = -3$ ... ( $\pi \approx 3.14$ , correct?)
• Jan 20th 2011, 06:09 PM
homeylova223
I might sound really dumb when I say this but what I do not really understand is how -pi and pi are asymptotes according to the back of my book because I usually think of them being zero.It boggle the mind because when I look at my unit circle I see pi as coordinates (-1,0) then I put opposite/adjacent which is zero why is it an asymptote here? That x/2 only changes period right or maybe there is something I am overlooking.
I mean I understand the post you put above me and this is what I really meant to ask. No disrespect.
• Jan 21st 2011, 05:03 AM
skeeter
Quote:

Originally Posted by homeylova223
I might sound really dumb when I say this but what I do not really understand is how -pi and pi are asymptotes according to the back of my book because I usually think of them being zero.It boggle the mind because when I look at my unit circle I see pi as coordinates (-1,0) then I put opposite/adjacent which is zero why is it an asymptote here? That x/2 only changes period right or maybe there is something I am overlooking.
I mean I understand the post you put above me and this is what I really meant to ask. No disrespect.

first note that $\tan\left(\pm \frac{\pi}{2}\right)$ is undefined.

so, for the function $y = \tan\left(\frac{x}{2}\right)$ ...

$\frac{x}{2} = \pm \frac{\pi}{2}$ at $x = \pm \pi$
• Jan 21st 2011, 11:47 AM
homeylova223
So what would that make -pi/2 and pi/2 as a domain what would you plug in the range. Would for -pi/2 would it be -2 and for pi/2 positive 2? Anyway I think I understand this now. And for x=zero y=zero I think I understand this now I was lost in darkness but now I see the light.