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Prove or contradict: Between each two solutions of $\arctan x = \sin x$ exists a solution for $1-\cos x = x^2 \cos x$

I have this question in a sample exam and I don't even know what would be a good way to approach this. I though about finding the ranges where the two difference functions have different slopes or something, but I'm not quite sure..

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    $\begingroup$ I don't know if you noticed it, or if it is even useful, but you can rearrange the second equation to get $\frac{1}{x^2+1}=\cos(x)$, and $\arctan'(x)=\frac{1}{1+x^2}$. $\endgroup$ – Botond Apr 24 '18 at 12:50
  • $\begingroup$ It's also useful here to note that $x=0$ is a solution for both equations. $\endgroup$ – Paul Apr 24 '18 at 12:56
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To prove it apply the Standard version of Rolle's theorem for $f\left( x \right)=\arctan \left( x \right)-\sin \left( x \right)$. Link

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  • $\begingroup$ Cool together with the hint from @Botond (which I didn't notice on my own) this is quite simple :) $\endgroup$ – Jason Apr 24 '18 at 13:15
  • $\begingroup$ you are welcome $\endgroup$ – user547564 Apr 24 '18 at 13:16
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we have ;

$1-\cos x = x^2 \cos x \implies \cos(x) = \frac1{x^2+1}$

integrate both sides ,

$\int\cos(x) \,dx = \int\frac1{x^2+1}\,dx$

$\sin(x) = \arctan(x) $ $\quad $

Note : i'm ignoring the constant ,because since $0$ is a solution the constants are equal and can be cancelled.

now $g(x) = \arctan(x)-\sin(x)=0$

Apply Rolles theorem,

since at roots the values are equal ie $0$, rolles theorem is applicable and proves that between each zeros of $\arctan(x)=\sin(x)$ there exists a root of $1-\cos(x) =x^2\cos(x)$

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  • $\begingroup$ Nice, I don't think the integration part of the proof is needed since it just follows from applying rolles theorem, doesn't it? $\endgroup$ – Jason Apr 24 '18 at 13:14
  • $\begingroup$ It isnt necessary, but i thought that point would show the connection between the two givens. $\endgroup$ – The Integrator Apr 24 '18 at 14:49

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