# How to graph a sin function with respect to it's limits and x intercepts

1. Mar 28, 2012

### Trespaser5

I have been given a question to sketch the curve of y=sin(x). I have looked into finding the domain which I understand but I don't understand how I prove the x intercepts mathematically as when I make x=0 I obviously get a 0 value for y but a sin curve obviously intercepts and pi and 2pi etc, how do I prove this ? Also why when I put sin(3pi/2) into the calculator do I not get a negative figure ? or why when I put pi and 2pi in do I not get zero ?

2. Mar 28, 2012

### LCKurtz

3. Mar 28, 2012

### Trespaser5

sorry, i didn't have my calculator in radian mode. Still I don't understand how I get the values of the x intercepts mathamatically in the same way as you do with polynomial expressions ?

4. Mar 28, 2012

### eptheta

It's actually part what periodic functions are.
for a periodic function g(x): g(x+P)=g(x) where P is the period.
So your function $f(x)=sin(x)$ is as good as $f(x)=sin (x \pm n\pi)$ n=0,1,2....
so $x=sin^{-1}(y) \pm n\pi$
now go ahead and put y=0 to get all your x intercepts.

5. Mar 28, 2012

### Trespaser5

that's the key to the door, thankyou :)

6. Mar 31, 2012

### Trespaser5

I've been trying all week but what I can't understand is that as [sin][/-1](0) is always 0 then any value I put in between 0 and 2∏ I get that value as an x intercept, where am I getting confused ?