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Mean value theorem for trigonometric function

  1. Feb 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Verify Lagrange's MVT for
    ## f(x)= sinx - sin2x ## in [ 0, π ]

    2. Relevant equations
    ## f'(c) = \frac{f(b)-f(a)}{b-a} ##

    3. The attempt at a solution
    Got on solving cosx= 2cos2x
    How to find c lies in [0, π ]?
    Solved it using quadratic equation but it gives a complicated value inside arccos.
     
    Last edited by a moderator: Feb 24, 2015
  2. jcsd
  3. Feb 24, 2015 #2
    On using that formula you get dy/dx at c = 0
    dy/dx = cosx-2cos2x now to verify MVT,dy/dx has to be equal to the value obtained from your equation for atleast one c within 0 to pi.
    putting cosx=2cos2x, you are indirectly finding the values of that c. If cosx can never be equal to 2cos2x in the given interval, then mvt is invalid here. Else MVT is valid.
     
  4. Feb 24, 2015 #3

    SammyS

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    Yes, the value inside the arccosine contains √(33) .
     
    Last edited by a moderator: Feb 24, 2015
  5. Feb 24, 2015 #4
    No it contains (1± √(33))/8.
    How to determine that it's in interval [0, π] ?
     
  6. Feb 24, 2015 #5

    SammyS

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    Well that does contain √(33) .

    arccos( y0) always gives a value in the interval [0, π] , provided that y0 is in the interval [-1, 1], the domain of the arccos function.
     
  7. Feb 24, 2015 #6
    But
    Cosx=2cos2x
    => cosx= 2( 2cos2x - 1)
    => cosx = 4cos2x - 2
    => cosx = (-1 ± √(33))/8
    So there is a contradiction with your statement?
     
  8. Feb 24, 2015 #7

    SammyS

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    I see √(33) right there in (-1 ± √(33))/8 .

    I just said it contains it.
     
  9. Feb 24, 2015 #8
    Sorry__ Sorry. I thought you meant that x= arccos(√(33)).
    My mistake.
    Thanks the major contribution from your side was telling me domain for arccos and range.
     
    Last edited: Feb 24, 2015
  10. Feb 24, 2015 #9

    Ray Vickson

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    Have you tried plotting the two functions over ##[0, \pi]##?
     
  11. Feb 24, 2015 #10
    See the plot
    What does that mean?
     
  12. Feb 24, 2015 #11

    Ray Vickson

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    Draw the graphs of ##y =2 \cos(2x)## and ##y = \cos(x)## for ##x \in [0,\pi]##, or perhaps the single graph of ## y = 2 \cos(2x) - \cos(x)##.
     
    Last edited: Feb 24, 2015
  13. Feb 24, 2015 #12
    Here it is newplot.
    Then what?
     
  14. Feb 24, 2015 #13

    Ray Vickson

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    Then look at it and think about what it is telling you.
     
  15. Feb 24, 2015 #14
    you use wolftam's graph calculator? Draw it yourself on a paper. Dont depend on graph calculators.
    see where the curves intesect. Those are the points where f'(c) is zero.
     
  16. Feb 25, 2015 #15
    But wolfram graph calculator gives correct results. I was using that here because it gives correct sketch and by hand we would get a rough sketch as drawing graphs of complicated functions could be difficult like
    sinx-sin2x.
    I think drawing only the graph of sinx-sin2x was okay?
    By seeing it the average slope is 0.
    Now according to MVT there is or are points on the function in given interval where the slopes of it equal average slope. Seeing the graph we see two points ( one is local minima and other is local maxima).
     
  17. Feb 25, 2015 #16
    By the way thanks Aditya, I saw 2 points where curves intersect and that means we get two values of c for f'(c) =0 .
    Also thank you Mr. Ray as I see two points where the curve cosx- 2cos2x cuts the x axis meaning we get two values of c for f'(c)=0.
     
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