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Trig problem

  1. Apr 14, 2008 #1

    I've been trying to solve the following but with no luck its so frustrating!

    if c= (n/pi) arccos [(n/pi)sin(pi/n)] + 2k(pi), where n is a positive integer,

    how can I show that c has precisely one solution in (0,1)?

  2. jcsd
  3. Apr 14, 2008 #2
    What have you done to try and solve it?

    Perhaps you can start by plugging in some positive integers for n...
  4. Apr 14, 2008 #3
    that won't be a complete proof tho, need to show it holds for all values of n. Have tried breaking it down and looking at the limits of certain parts within the function, e.g sin (pi/n) lies between 0 and 1 for all n etc but this doesn't seem to get me anywhere as I get stuck when i reach the (n/pi) in front of the arccos!
  5. Apr 14, 2008 #4
    You are right, it won't be a complete proof. But by trying and working out the first few integer values, you will have a better understanding as to why it will always be between 0 and 1. Then perhaps you can produce a proof by induction.
  6. Apr 14, 2008 #5
    i know that it lies between 0 and 1, but it is trying to show that there is only ONE solution between 0 and 1 that i haven't been able to achieve.I don't think it is possible by induction or contradiction but that a theorem or consequences of a theorem is required :confused:
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