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Inequality - how to prove that xcosx < sinx for all x

  1. Apr 20, 2008 #1
    Hello!! does anyone know how to prove that xcosx < sinx for all x greater than 0?

    Many thanks
     
  2. jcsd
  3. Apr 20, 2008 #2

    quasar987

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    Well it's obviously false, for take x=2pi.
     
  4. Apr 20, 2008 #3
    sorry my error, the interval for x should be 0 < x < (pi)

    sorry!
     
  5. Apr 20, 2008 #4

    quasar987

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    Well, notice that the problem is equivalent to showing that xcotx<1 for all x in (0,pi). So use calculus.
     
  6. Apr 20, 2008 #5
    [tex] tan(x)>x=>\frac{sin(x)}{cos(x)}>x=>sin(x)>xcos(x)[/tex] THis is valid however on the interval [tex]\left(0,\frac{\pi}{2}\right)[/tex]

    Edit: Or better use quasar987's suggestion, it is better and valid on the whole interval.
     
    Last edited: Apr 20, 2008
  7. Apr 20, 2008 #6
    thanks guys, but im failing to see how its easier to solve the inequality when written as xcotx<1? :confused:
     
  8. Apr 20, 2008 #7

    quasar987

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    Do you know calculus? (differentiation and things like that)
     
  9. Apr 20, 2008 #8
    yes thanks, have been using differential calculus to try and solve the problem I posted but have not been able to, else I wouldn't have posted it! just dont see the advantages of changing to xcotx<1
     
  10. Apr 20, 2008 #9
    Consider the function f(x) = xcos(x)-sin(x). If you know calculus it should be fairly easy to show where this function is negative!
     
  11. Apr 20, 2008 #10
    Again, this is what I have been trying to do! are we for example taking f(x), noting that f(0) =0, and using the first derivative test to show that the function is decreasing ?
     
  12. Apr 20, 2008 #11
    ok, here it is how i would argue about showing the validity of that inequality using calculus.

    Lets define the following function:

    [tex] f(x)=xcosx-sinx,\forall x \in \left(0,\pi \right)[/tex]

    NOw,

    [tex]f'(x)=cosx-xsinx-cosx=-xsinx[/tex]

    Now let's find the critical numbers

    [tex] f'(x)=0=>xsinx=0=>x=0,and,x=k\pi[/tex]

    But, notice that none of these numbers are in the domain of our function, so it means that f, has no cr numbers at all.

    THis actually means also that f'(x) does not change sign at all on the given interval, so it is sufficient to plug in any value from the interval (0,pi) to determine the sign of f' on the whole interval, that is let [tex]x=\frac{\pi}{2}[/tex]

    SO:

    [tex]f'(x)=\frac{\pi}{2}sin\frac{\pi}{2}=-\frac{\pi}{2}<0[/tex]

    THis means that our function is decreasing on the given interval. Now from this information we know that:

    [tex] f(0)>f(0+\epsilon),\epsilon>0,\epsilon\to\ 0, \epsilon \in (0,\pi][/tex]


    Hence,

    [tex]f(0)=0*cos(0)-sin(0)=0[/tex] it means that

    [tex] f(x)<0,\forall x \in (0,\pi)[/tex]

    so:

    [tex]f(x)=xcosx-sinx<0=>xcosx<sinx[/tex]
     
  13. Apr 20, 2008 #12
    ahh thank you very much stupidmath! Also, when solving inequalities like this in the future, should i always try to solve for critical nos, or is it enough to say for example in this case that since f(0)=0 and f'(x) = -xsinx, which is clearly negative over the whole the interval then f(x) is a decreasing function and hence less than 0?
     
  14. Apr 20, 2008 #13
    Take [tex] x=\frac{3 \pi}{2}[/tex] f' is clearly >0 for this value. Or take any x in the interval (pi,2pi) what do you get for f'=-xsinx ??????.SO it means that our funcitno is decreasing only on (0,pi) but this does not guarantee us that it is decreasing everywhere in its domain, like in this case which is not ture, since f is not decreasing everywhere but rather only on (0,pi) for our purporse.

    So actually finding cr. points and seeing where function is increasing and decreasing guarantees us where f' changes sign and where not.
     
    Last edited: Apr 20, 2008
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