# Show that $\exists x \in (0, \pi/2)$ such that $\cos(x)=x$

Let $f(x) = \cos(x)$, $g(x)=x$, both functions are continuous. $f(0)=1$, $f(\pi/2)=0$, so, by the Intermediate Value Theorem, for any $z\in [0,1]$, there exists $c\in [0,\pi/2]$ such that $f(x)=z$.

This should be simple to prove, but for some reason I have a problem with IVT, don't know why. Would appreciate some help.

I think that the fact that $g(x)=x$ is onto $[0,1]\subset [0, \pi/2]$ should be used here somehow.

• Subtract them, then IVT. Commented Nov 18, 2016 at 4:30

Consider $h(x)=\cos(x)-x$. Then, $h(0)=1$ and $h\left( \displaystyle\frac{\pi}{2}\right)=-\displaystyle\frac{\pi}{2}$. Hence, $h(0)>h\left( \displaystyle\frac{\pi}{2}\right)$ and, clearly, $h(x)$ is continuous. For the IVT, there exist a $z\in\left( 0,\displaystyle\frac{\pi}{2}\right)$ such that $h(z)=0$.

$h(z)=0$ if and only if $\cos(z)-z=0$ if and only if $\cos(z)=z$

• You could also use $h(1)=\cos(1)-1<1-1=0$ to get the argument working on the smaller interval. Commented Nov 18, 2016 at 13:08

If a solution for $$\cos(x) = x$$ exists in $$[0, \pi/2]$$, it must lie somewhere in the interval $$[0, 1]$$ as $$\cos(x)$$ is bounded above by $$1$$. If we can show that $$\cos:[0, 1] \rightarrow [0, 1]$$ is a contraction mapping, then the Banach fixed-point theorem guarantees a fixed point on this interval—that is, a point where $$\cos(x) = x$$. By definition, it is a contraction mapping if there exists a $$q \in [0, 1)$$ such that $$|\cos(x) - \cos(y)| \leq q|x-y|$$ for all $$x, y \in [0, 1]$$.

Rearranging this yields $$\displaystyle \frac{|\cos(x) - \cos(y)|}{|x-y|} \leq q$$. How large can the left hand side get? The mean value theorem tells us the left hand side will be bounded by the absolute value of the largest possible evaluation of cosine's derivative on $$[0, 1]$$. We have $$\displaystyle \frac{\operatorname{d}}{\operatorname{dx}} \cos(x) = -\sin(x)$$. Because $$\sin(0) = 0$$, and because $$\sin$$ is strictly increasing on $$[0, \pi/2]$$, we can safely conclude that the left hand side of the above inequality is bounded above by $$\sin(1)$$. Since $$0 < \sin(1) < \sin(\pi/2) = 1$$, this value can serve as our $$q$$.

Therefore, cosine on the interval $$[0,1]$$ is a contraction mapping $$\implies$$ a fixed point is guaranteed by the Banach fixed-point theorem.

• An advanced answer for an elementary question, nevertheless great Commented Nov 18, 2016 at 5:15

Define $h(x)=\cos x-x$

Then $h(0)=1>0;h(\frac{\pi}{2})=-\frac{\pi}{2}<0$

By IVP $\exists c\in (0,\frac{\pi}{2})$ such that $h(c)=0$

The proof is immediate by the use of graphs of $$y=\cos x$$ and $$y=x$$ in the given interval. See this Wolfram Alpha link for plots