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Find absolute extrma of sin(cos(x))

  1. Mar 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A simple question; find the absolute extrema of f(x)=sin(cos(x)) on the interval [0,2pi].


    2. Relevant equations
    The chain rule


    3. The attempt at a solution
    Assuming that I am correct about the derivative being -cos(cos(x))sin(x), how do you solve
    -cos(cos(x))sin(x)=0
     
  2. jcsd
  3. Mar 25, 2010 #2

    Borek

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    Assuming you are right - when does x*y equal 0?
     
  4. Mar 25, 2010 #3
    The absolute extrema must occur at the endpoints of the interval or at the solutions to the equation f'(x)=0. Do I change the trigonometric product into a sum using sin(A+B)+sin(A-B). When is x*y=0? when either x or y =0.
     
  5. Mar 25, 2010 #4

    Borek

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    Can't you use this information to solve your equation?
     
  6. Mar 25, 2010 #5
    So you are saying that sin(x)=0 => x=0,pi,2pi and how about the cos(cos(x))=0.
     
  7. Mar 25, 2010 #6
    for what t's cos(t)=0?
     
  8. Mar 25, 2010 #7
    For what t's cos(t)=0, t= pi/2, 3pi/2
     
  9. Mar 27, 2010 #8

    Borek

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    What values can cos(x) take? If so, what values can cos(cos(x)) take?
     
  10. Mar 27, 2010 #9
    the cos(x) can take on all values in the interval [-1,1], depending on the value of x.
     
  11. Mar 27, 2010 #10

    Borek

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    Take it a step further - can you tell anything about cos([-1,1])?

    And start moving on your own, my hand hurts from spoonfeeding.
     
  12. Mar 27, 2010 #11
    I know all that, but I can't see it as a solution. Obviously cos([-1,1]) is positive only. I hope your hand doesn't hurt too much. Thanks very much for the help.
     
  13. Mar 27, 2010 #12

    Borek

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    x*y=0 if y is always > 0.
     
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