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To maximise a trigonometric function

  1. May 12, 2009 #1
    1. The problem statement, all variables and given/known data

    What is the maximal value of
    f(x) = sin(x) - [tex]\sqrt{3} cos(x)[/tex]
    such that
    0 <= x < 360?

    3. The attempt at a solution

    To raise to power to two

    [tex] f^2 = sin^2(x) - 2 \sqrt{3} sinx cosx + 3cos^2(x) [/tex]

    I obverse that the last term has the greatest coeffient so
    I maximise it by
    setting x = 0.

    We get
    f^2(0) = 3
    and

    f = 3^.5

    However, the right answer is 2.

    How can you get the right answer?
     
  2. jcsd
  3. May 12, 2009 #2

    Cyosis

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    Homework Helper

    Why did you take the square? To maximize the function you need to take the derivative of f with respect to x and equal it to zero. You will find two types of x-es, one that minimizes and one that maximizes. Start with calculating f'(x).
     
  4. May 12, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Another way to do that, without differentiating, is to use the trig identity sin(r+ t)= sin(r)cos(t)+ cos(r)sin(t). Of course, you can't find t such that sin(t)= 1 and [itex]cos(t)= \sqrt{3}[/itex] because you must have [itex]cos^2(t)+ sin^2(t)= 1[/itex] while
    [tex]1^2+ (\sqrt{3})^2}= 1+ 3= 4[/itex]
    Okay, write that as
    [tex]2(\frac{1}{2}sin(x)+(-\frac{\sqrt{3}}{2}cos(x))[/itex]

    Then [itex]sin(x)+ \sqrt{3}cos(x)= 2 sin(x+ \theta)[/itex] where [itex]\theta[/itex] is the angle such that [itex]cos(\theta)= 1/2[/itex] and [itex]cos(\theta)= \sqrt{3}/2[/itex]. That's easy to find, but it really doesn't matter what [itex]\theta[/itex] is.
     
  5. May 12, 2009 #4
    Thank you for your answer!

    I really remember that to raise to the power 2 is often useful in trig calculations.
    It seems that this time the strategy:
    1. raise to the power 2
    2. and consider the extreme situation
    did not work.

    I believe that you can get the same result also with my strategy.
    However, I am unsure how to get 2 out of my formula.
    The best which I can do is
    3^.5
     
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