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Using the mean value theorem on trig functions

  1. Jul 27, 2011 #1
    1. The problem statement, all variables and given/known data

    let g be a function mapping x to xcosx-sinx.
    use the mean value theorem to prove that g(x) < 0 for x in (0,pi]


    2. Relevant equations

    well the function is both continuous and differentiable on the interval so that's a start...



    3. The attempt at a solution

    basically i thought i'd separate the interval into smaller sections coinciding with the roots of xcosx and sinx, ie:

    i ticked off the interval [pi/2,pi] as xcosx is negative and sinx is still positive so g(x) <0;

    currently considering xcosx for x in (0,1) and trying to find a turning point but i'm not seeing anywhere i can actually apply the MVT at the moment... halp!
     
    Last edited: Jul 27, 2011
  2. jcsd
  3. Jul 27, 2011 #2

    micromass

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    Hi magik heepo! :smile:

    The trick is to show that the derivative of g is always negative on [0,pi]. You should apply the mean value theorem to get that g is decreasing.
     
  4. Jul 27, 2011 #3
    Thanks for the reply micromass!

    So now I write out the mvt as

    ((acos(a)-sin(a))-(bcos(b)-sin(b)))/a-b = -csin(c) and state that

    since sin(c)>0 for c in [0,pi] and c>0 so -c<0. Also we've set a>b so a-b is positive, leaving

    (acos(a)-sin(a))-(bcos(b)-sin(b)) = (b-a)csin(c) < 0

    so g(x) <0...

    well i'm convinced, thanks again :)
     
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