# Prove that the equation $\cos x - x -1/2 = 0$ has a unique real solution

Prove that the equation $\cos x - x -1/2 = 0$ has a unique real solution.

My solution: I was starting with a function $F(x) = \cos x - 1/2$ and the interval $[0,\pi/4]$ and trying to show that the fixed point theorem is applicable here. If I can find this solution to a few decimal places then I have shown that there is unique real solution. However, I am having some trouble applying the Fixed Point Theorem to $F(x)$. Any help would be greatly appreciated.

• There are quite a few fixed point theorems, which one are you using? There is one where, if you can show that $F(x)$ is a contraction map, that you will indeed have a fixed point. Don't forget about the power of the Intermediate Value Theorem here as well. – DaveNine Sep 24 '15 at 15:06
• Where did you get that exercise? – Nash Nov 16 '16 at 16:18

An other way

Let $f(x)=\cos(x)-\frac{1}{2}$. $$|f(x)-f(y)|=|\cos(x)-\cos(y)|.$$ Using the mean value theorem, we can show that $$|\cos(x)-\cos(y)|\leq \frac{\sqrt 2}{2}|x-y|$$

and thus $f$ is a contraction mapping. By Banach fixe point theorem, $f$ has a unique fix point, and thus, there is a unique $x$ such that $f(x)-x=0$ what prove the claim.

By mean value theorem, there is a $c=c_{x,y}\in]0,\pi/4[$ such that $$\cos(y)-\cos(x)=-\sin(c)(y-x)\implies -|\sin(c)||y-x|\leq \cos(y)-\cos(x)\leq |\sin(c)||y-x|\implies |\cos(x)-\cos(y)|\leq |\sin(c)||y-x|\leq \sin(\pi/4)|y-x|=\frac{\sqrt 2}{2}|y-x|.$$

• Can you edit to show how you got <= √2/2 |x−y| – YesItsMario Sep 24 '15 at 15:29
• Ok, it's done :-) – idm Sep 24 '15 at 15:37

Let $f(x)=\cos x-x-1/2$. $f'(x)\le0$ implies $f$ is monotonic. And the zeros of $f'$ are isolated, so $f$ is strictly monotonic. Note that $f(0)>0$ and $f(1000)<0$.

• This proves there are some zeroes, not that there is a unique zero. – Did Sep 25 '15 at 17:46
• @Did Thanks. I have filled that gap. – Eclipse Sun Sep 26 '15 at 0:39

$F(x) = \cos x - \frac{1}{2}$ then $F'(x) =-\sin x \Rightarrow \vert -\sin x \vert \leq \frac{\sqrt{2}}{2}<1$ and $F'(x) =-\sin x<0$ in $[0,\frac{ \pi}{4}]$ thus max happen in $0$ and min in $\frac{ \pi}{4}$ where $f(0)=\frac{1}{2}$ and $f(\frac{ pi}{4})=\frac{\sqrt{2}}{2}-\frac{1}{2}$ thus $F$ maps $[0,\frac{ \pi}{4}]$ to $[0,\frac{ \pi}{4}]$. Now apply the Fixed Point Theorem to F(x). finished

Another approach to this problem can be:

Let $f(x)=\cos x$ and $g(x)=x+\frac{1}{2}$

The first and the most important thing we note here is that these two functions are continuous

Now, at $x=0$ $$f(0)=1>\frac{1}{2}=g(0)$$ At $x=π/2$ $$f(π/2)=0<\frac{π+1}{2} =g(π/2)$$

Thus, there has to exist a real point P where these two points meet, i.e. at some real $x=u$, $$f(u)=g(u)$$ $$\implies \cos u=\frac{1}{2}+u$$ $$\implies \cos u-u-\frac{1}{2}=0$$ And thus we have proved the existence of a real solution for this equation. $$Q.E.D.$$