# Graphing csc and cot functions?

• January 17th 2011, 12:54 PM
homeylova223
Graphing csc and cot functions?
I am having trouble graphing these functions
y=cot(1-pi/2)
I know the period is pi
the phase shift is pi/2
But I cannot understand how to graph it

the second function I am having trouble graphing is
y=csc(2+pi)-3
I know the period is pi
the phase shift is -pi/2
and the vertical shift is -3
But I do not understand how to graph it either

• January 17th 2011, 01:10 PM
pickslides
Quote:

Originally Posted by homeylova223
I am having trouble graphing these functions
y=cot(1-pi/2)

y=csc(2+pi)-3

Are these functions? Maybe you have left out a $\theta$ or an $x$ ?
• January 17th 2011, 01:16 PM
homeylova223
Yes I did not put the symbol you typed before x because I am not sure how to make that symbol with my keyboard that symbol you typed that look like an o I am not sure what it is called. That is the symbol that goes next to my 1 and 2 in the function sorry.
• January 17th 2011, 01:26 PM
pickslides
Quote:

Originally Posted by homeylova223
Yes I did not put the symbol you typed before x because I am not sure how to make that symbol with my keyboard that symbol you typed that look like an o I am not sure what it is called. That is the symbol that goes next to my 1 and 2 in the function sorry.

That symbol is called "theta" it has Greek origins, very popular in Mathematics when describing angles.

$\displaystyle y=\cot\left(\theta -\frac{\pi}{2}\right)$

$\displaystyle y=\csc\left(2\theta +\pi\right)-3$

Do you know the general shape of these functions?

Are you aware that

$\displaystyle y=\cot\left(\theta -\frac{\pi}{2}\right) =\frac{1}{\tan\left(\theta -\frac{\pi}{2}\right)}$

$\displaystyle y=\csc\left(2\theta +\pi\right)-3=\frac{1}{\sin\left(2\theta +\pi\right)}-3$
• January 17th 2011, 04:27 PM
homeylova223
Ah I see the trouble I am having is that I am not sure how to use the phase shift on the cot function. I know the asymptotes of a regular cot function the only thing I am not sure is how to apply the phase shift. For example on the first one I know the phase shift is pi/2 and the period is pi. So would I start at pi/2 and end at 3pi/2 in order to complete the function? The thing is that the cot of pi/2 is zero then the cot of the next one 3pi/4 is negative 1 so I am kind of confused on how to make the phase shift.
• January 17th 2011, 04:46 PM
pickslides
Quote:

Originally Posted by homeylova223
The thing is that the cot of pi/2 is zero

It is, but you take $\frac{\pi}{2}$ within the function.

$\displaystyle \theta = \frac{\pi}{2} \implies y=\cot\left(\frac{\pi}{2} -\frac{\pi}{2}\right) =\frac{1}{\tan\left(\frac{\pi}{2} -\frac{\pi}{2}\right)}$

is now the asymptote.
• January 17th 2011, 06:10 PM
homeylova223
Oh I see so I have to plug in my values like pi/2 and 3pi/2 into the function. Then I graph these value ok.