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

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..

• 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 $\arctan'(x)=\frac{1}{1+x^2}$. – Botond Apr 24 '18 at 12:50
• It's also useful here to note that $x=0$ is a solution for both equations. – Paul Apr 24 '18 at 12:56

## 2 Answers

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

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

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 $1-\cos(x) =x^2\cos(x)$

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