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Calc Help

  1. Apr 18, 2005 #1

    This is a fairly straightforward problem but I want to do it using calculus. Here goes:

    Prove that [itex]\sin 1 > \cos(\sin 1)[/itex].

    This is what I've done (I've hardly done much):

    Let [itex]f(x) = \sin(\cos(\sin x) - \cos(\sin(\cos(x)))[/itex]. I also have to show that f(x) = 0 has exactly one solution in [itex][0,\pi/2][/itex]. So anyway, f'(x) < 0 for all x in this interval which in particular means that f(0) > f([itex]\pi/2[/itex]). But the first part requires showing that f(0) > 0. Can this be done by a purely calculus-based argument? It is easily done if we bend the inequality a bit and observe that it is true if [itex]\pi/2 -1 <\sin 1[/itex].

    Thanks and cheers
  2. jcsd
  3. Apr 21, 2005 #2
    Hi all

    This was buried deep inside the forum and I had to fish it out. Just a reminder asking for help with it...

    Thanks and cheers
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