Show that there is some $x \in (0,\frac{\pi}{2})$ such that $\cos(x) = x$ and prove this is the only real solution.

The first part I know how to do, so quickly in case somebody does look this up:

Let $f(x) = \cos(x) - x$. Examining $f(x)$ at the points $0$ and $\frac{\pi}{2}$ we obtain: $$f(0) = 1 \ and\ f(\frac{\pi}{2}) = -\frac{\pi}{2}$$.

By IVT there exists a point $x \in (0,\frac{\pi}{2})$ such that $f(x) = 0$. $$\therefore\ \cos(x) - x = 0 \ \Rightarrow \ \cos(x) = x$$

So this being the case to show this is the only real solution, suppose not. Then there are $x_{1}$ and $x_{2}$ such that: $\cos(x_1) = x_1$ and $\cos(x_2) = x_2$. Performing some algebra I arrive at the following: $$\cos(x_1) - \cos(x_2) = x_1 - x_2$$.

I tried using a trig formula, but the only one that applied produced this: $$-2\sin\Bigg(\frac{x_{1} + x_{2}}{2}\Bigg)\sin\Bigg(\frac{x_{1} - x_{2}}{2}\Bigg) = x_{1} - x_{2} $$

I don't see this helping me in any way. Is there another form of showing this ?

  • 3
    $\begingroup$ Try checking the sign of the derivative; you can prove the function is strictly monotone guaranteeing that the solution is unique. $\endgroup$ – Clayton Jun 24 '19 at 22:25
  • $\begingroup$ I know I should know this theorem, but I don't remember it. Where could I find it? $\endgroup$ – dc3rd Jun 24 '19 at 22:30
  • $\begingroup$ It's a basic property of scalar differentiable functions. Check any calculus notes $\endgroup$ – Gonzalo Benavides Jun 24 '19 at 22:48
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    $\begingroup$ If a function is strictly monotone on an interval there are by definition no two equal values of the function on the interval. Particularly there are no two (or more) zeros. $\endgroup$ – user Jun 24 '19 at 22:50

Let's look at $f(x)=x-cos(x)$.

The way you showed that there is at least one solution is technically correct, however you need to explain why using the IVT is allowed (In this case you need to explain why $f(x)$ is continuous in $(0,\pi/2)$).

Now, let's assume that there are two different points such that $x=cos(x)$ in $(0,\pi/2)$. We will call them $x_1$ and $x_2$, and we will say that $x_1<x_2$.

That means that $f(x_1)=f(x_2)=0$.

Applying Rolle's theorem in $[x_1,x_2]$ (Do you understand why it may be applied here?), we get that there is a point $x_1<x_0<x_2$ such that $f'(x_0)=1+sin(x_0)=0$.

However, there is no such point in the interval $(0,\pi/2)$, so there is a single solution, because more than a single solution is impossible, and because you showed that a solution exists.

EDIT: I misread the question. Assume that there is another solution $x_1\in \mathbb{R}$, such that $x_1>x_0$ where $x_0$ is the solution you have found in the domain $(0,\pi/2)$. I have already shown that no other solution exists in $(0,\pi/2)$.

$f(\pi/2)=\pi/2$, and $f(x_1)=0$. But $x_1>\pi/2$, so there must be an interval $I\subset [pi/2,\infty]$ where $f$ is decreasing.

Because $f$ is differentable in $\mathbb{R}$, if $f$ is decreasing in an interval, there must be at least one point in it such that $f'(x)<0$, but that is impossible for $x\geq\pi/2$, so there is no such solution.

Showing that there are no solutions for $x_1<x_0$ is very similar.

  • $\begingroup$ The question also asks to show there are no other real solutions, not just uniqueness of solution in $(0, \pi/2)$. To this end: if $\cos x = x$, then $-1 \le \cos x \le 1$ so $-1 \le x \le 1$; and then since $\cos x > 0$ whenever $-1 \le x \le 1$ it also follows that $x > 0$. $\endgroup$ – Daniel Schepler Jun 24 '19 at 23:04

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