Graphing Trigonometric Functions Question

[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?

 

[Integral]

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

 

[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.

 

[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.

 

[Oggy]

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