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Using the mean value theorem, prove that, for $0<x<\pi/2$, $\cos x<\cos(\sin x)$.

I was trying to use the mean value theorem but I got lost. I am a newbie to this please explain this as you're explaining it to your worst student!

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    $\begingroup$ Use $x>\sin x$. $\endgroup$ – Lord Shark the Unknown May 18 '18 at 6:24
  • $\begingroup$ as Dark Mathorp pointed that you might need to show that $x>\sin(x)$ so I show you how to show that $x>\sin(x)$. Note that $x-\sin(x)$ is a convex function for $x=(0,\pi /2)$ and its minimum value is obtained when $x=0$. Hence $x>\sin(x)$ for your desired values of $x$. $\endgroup$ – Frank Moses May 18 '18 at 6:29
  • $\begingroup$ Okay I understood your answer but is the proof doable using mean value theorem? $\endgroup$ – aakash anand May 18 '18 at 6:31
  • $\begingroup$ I took f(x) = cos x - cos(sin x) and I'm trying to prove that this function is strictly decreasing $\endgroup$ – aakash anand May 18 '18 at 6:39
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HINT

Note that

  • $x>\sin x$
  • $\cos x$ is a strictly decreasing function on the interval
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You can show that $\cos(x)$ is decreasing on that and then use the fact that $x > \sin(x)$ (you may have to show this as well, but that is not a challenge).

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