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Maximum of a trigonometric function

  1. Apr 29, 2010 #1
    This is related to my previous https://www.physicsforums.com/showthread.php?t=398964". I am having trouble to get the maximum of the following trigonometric function:

    [tex]

    \cos^m\theta_1(c\cos^n\theta_2+s\sin^n\theta_2)+\sin^m\theta_1 ( c\sin^n\theta_2+s\cos^n\theta_2)

    [/tex]

    Here [tex]m,n\ge2 \mbox{ are fixed positive integers and } c,s[/tex] are fixed positive reals with [tex]c^2+s^2=1[/tex]. The maximum is to be carried out w.r.t. [tex]\theta_1,\theta_2[/tex] in the range [tex]0\le\theta_1,\theta_2\le\frac{\pi}{2}[/tex] In my trying, I got the maximum to be max(c,s), but I fear may be I have done some mistake. I got the result by differentiating w.r.t. [tex]\theta_1,\theta_2[/tex] and vanishing them....but I have ignored the case when [tex]\cos\theta_1\cos\theta_2\sin\theta_1\sin\theta_2\ne0[/tex]. Can anybody help me, please.

    Can I say that the function is all time differentiable within its closed and compact domain (the rectangle), the maximum should be attained on boundary?
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Apr 29, 2010 #2
    Plotting in mathematica with different values, I can not find an exception.

    But the reasoning of closed, compact domain can not be applied, because though the domain is convex, the function is itself not a convex one. However, we can split the domain such that in each part, the function remains monotonic.
     
  4. Apr 29, 2010 #3
    Oh, finally I got an answer to this question...the proof is a handy one (according to me, of course!:smile:). The answer is indeed correct.

    Since [tex]\cos\theta_1\cos\theta_2\sin\theta_1\sin\theta_2\ne0, \cos\theta_1,\cos\theta_2,\sin\theta_1,\sin\theta_2<1.[/tex]
    so

    [tex]\cos^m\theta_1(c\cos^n\theta_2+s\sin^n\theta_2)+
    \sin^m\theta_1 ( c\sin^n\theta_2+s\cos^n\theta_2)[/tex]

    [tex]\le
    \cos^2\theta_1(c\cos^2\theta_2+s\sin^2\theta_2)+
    \sin^2\theta_1 ( c\sin^2\theta_2+s\cos^2\theta_2)[/tex]

    [tex]\le\frac{c+s}{2}[/tex]

    [tex]<\max\{c,s\} [/tex]

    Regards.
     
    Last edited: Apr 29, 2010
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