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

  1. Mar 28, 2012 #1
    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. jcsd
  3. Mar 28, 2012 #2

    LCKurtz

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    Check that your calculator is in radian mode, not degree mode.
     
  4. Mar 28, 2012 #3
    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 ?
     
  5. Mar 28, 2012 #4
    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 [itex] f(x)=sin(x) [/itex] is as good as [itex] f(x)=sin (x \pm n\pi) [/itex] n=0,1,2....
    so [itex] x=sin^{-1}(y) \pm n\pi [/itex]
    now go ahead and put y=0 to get all your x intercepts.
     
  6. Mar 28, 2012 #5
    that's the key to the door, thankyou :)
     
  7. Mar 31, 2012 #6
    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 ?
     
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