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Trigonometric function with specific properties

  1. May 14, 2009 #1
    Is there any trigonometric function f(theta) where theta is an angle such that

    f(Pi-theta)=-f(theta) and such that its square (i.e f(theta)^2) is maximum at Pi/2?


    Thanks
     
  2. jcsd
  3. May 14, 2009 #2
    try theta = PI/2

    then f(PI-(PI/2)) = f(PI/2)

    so:

    if f(PI-theta)=-f(theta) for all theta, then f(PI/2) must equal zero.

    and you also said that:

    f(PI/2)^2 is its maximum. but f(PI/2)^2 is 0.

    so f(theta) = zero, for all theta. A boring and useless function.

    edit: Presuming f(theta) cannot be infinite
     
    Last edited: May 14, 2009
  4. May 14, 2009 #3

    tiny-tim

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    Hi Physicslad78! Hi Georgepowell! :smile:

    (have a pi: π and a theta: θ :wink:)
    It depends exacty what the question envisages …

    if f is allowed to be infinite at π/2, then you could have f(π/2)- = -f(π/2)+ = ±∞ :wink:
     
  5. May 14, 2009 #4
    Ahh, I missed that. How about the tan() function then?
     
  6. May 14, 2009 #5
    Thanks a lot guys..f cannot be infinite as it is a wave function in a physics system I am researching on..I just wanted to make sure that there aint any solutions except the f(theta)=0 as tiny-tim pointed out..Thanks again...
     
  7. May 14, 2009 #6
    Wait... what about sin(x)?

    sin(Pi-theta) = sin(PI)cos(theta) - sin(theta)cos(pi) = -sin(theta)

    And [sin(x)]^2 has local maxima at PI/2 + nPI.

    Am I missing something?
     
  8. May 14, 2009 #7

    tiny-tim

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    Yup! … cosπ = minus 1 :wink:
     
  9. May 14, 2009 #8
    oops. lol.
     
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