# Graphing Trigonometric Functions Question

1. Feb 21, 2005

### DLxX

Ok I can graph sin(x) , cos(x) , and tan(x) pretty easily, but I'm having a hard time graphing the csc, sec, and cot ones. For the first three I just found the values of pi/2 pi 3pi/2 and 2pi. So for example pi/2 for sin(x) would be sin 90 or sin pi/2 which is equal to 1. I then just put a dot at 1 and did the same thing for the rest of the angles in radian measure and then just draw a smooth curve through them, but how do I do this with csc, sec, and cot?

2. Feb 21, 2005

### Integral

Staff Emeritus
Consider what happens to csc($\Theta$) When Sin($\Theta$)=0 .

3. Feb 22, 2005

### HallsofIvy

In other words, you can't draw "smooth" graphs for sec(x) and csc(x) (or for that matter tan(x) and cot(x)). They aren't "smooth"- they aren't even continuous.

4. Feb 22, 2005

### dextercioby

You should have seen that when u graphed "tan".You said u graphed it,so u have noticed that is "blows up" in certain points (namely where the cosine is zero).It's discontinuous as well.

Daniel.

5. Feb 22, 2005

### Oggy

$$\tan{x}$$, 'breaks up' at $$\frac{\pi}{2}(2k-1)$$, and $$\cot{x}$$, 'breaks up' at $$k\pi$$.