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Can someone help me prove that $0 \leq 1 -\cos(x) \leq x^2/2$

The hint from the book, was that I could assume that $ x \in (0, PI/2), $ then I could use Pythagoras' to realise that $ |QP| = \sqrt(2 -2cosx)$ and use that the line QP must be shorter than the arc lenght between the two points. And from there I can use the same arguemnt for $ x \in (Pi/2, PI)$. But I don't quite understand what to do.

enter image description here

tonytouch
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    What can you use? – José Carlos Santos Nov 06 '17 at 08:39
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    Can you use Taylor series? If not, first consider that for angles in the 1st quadrant, $\sin x \le x \le \tan x$ – PM 2Ring Nov 06 '17 at 08:45
  • The hint from the book, was that I could assume that $ x \in (0, PI/2), $ then I could use Pythagoras' to realise that $ |QP| = \sqrt(2 -2cosx)$ and use that the line QP must be shorter than the arc lenght between the two points. And from there I can use the same arguemnt for $ x \in (Pi/2, PI)$. But I don't quite understand what to do. – tonytouch Nov 06 '17 at 08:48

5 Answers5

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Based on what you have written in comments about your textbooks idea: consider the unit circle with center at $O$ and points $P$ and $Q$ on the circle such that $\angle POQ=x$. Let us assume that $x \in (0, \pi/2)$. enter image description here

Consider the $\triangle POQ$, let $OM$ be the perpendicular from $O$ onto the line $PQ$. In $\triangle OMP$, we have $$\sin \left(\frac{x}{2}\right)=\frac{MP}{OP}=\frac{MP}{1}=MP.$$ Thus the segment $PQ=2MP=2\sin \left(\frac{x}{2}\right)$.

Since the arc $PQ$ is greater than or equal to the line segment $PQ$, therefore we have $$\text{segment }PQ = 2\sin \left(\frac{x}{2}\right) \leq x=\text{arc }PQ.$$ This implies \begin{align*} 2\sin \left(\frac{x}{2}\right) &\leq x\\ 2\sin^2 \left(\frac{x}{2}\right) &\leq \frac{x^2}{2}\\ 1-\cos x & \leq \frac{x^2}{2}. \end{align*}

Anurag A
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$\cos x\le1$ can be taken for granted. Then if we integrate twice from $0$ to $x$,

$$\int_0^x\cos x\,dx=\sin x\le x,$$

$$\int_0^x\sin x\,dx=1-\cos x\le\frac{x^2}2.$$

Note that we can continue forever

$$\int_0^x(1-\cos x)\,dx=x-\sin x\le\frac{x^3}6,$$

$$\int_0^x(x-\sin x)\,dx=-1+\frac{x^2}2+\cos x\le\frac{x^4}{24},$$

$$\cdots$$

which leads us to the Taylor development.

Successive upper and lower polynomial bounds of the cosine:

enter image description here

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The inequality $\cos(x)\leq 1$, is trivial because $\cos(x)\leq \sqrt{\cos^2(x)+\sin^2(x)}=1$.

For the second one let $f(x):=x^2/2-1 +\cos(x)$. We have to show that $f(x)\geq 0$. Since $f(-x)=f(x)$, it suffices to consider the case when $x\ge 0$. Now $$f'(x)=x-\sin(x)\quad\text{and}\quad f''(x)=1-\cos(x)\geq 0$$ which means that $f'(x)$ is increasing and $f'(x)\geq f'(0)=0$. Hence also $f$ is increasing for $x\geq 0$, that is $f(x)\geq f(0)=0$.

Robert Z
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Because of symmetry it suffices to consider non-negative values of $x$. $$1-\cos(x)=\int_0^x\sin(y)\,{\rm d}y\leq\int_0^xy\,{\rm d}y=x^2/2.$$

Gerhard S.
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    You are silently using $y\le\sin y$ without justification. –  Nov 06 '17 at 08:53
  • I like this method ! (+1) @YvesDaoust : Just a draw is enough to justify that $|\sin(y)|\leq |y|$ (and not the inverse inequality as you wrote). It's moreover a very very known inequality... – idm Nov 06 '17 at 09:01
  • @idm: "It's moreover a very very know inequality" is a very poor argument. Because $\cos x\ge 1-x^2/2$ is also a very very known inequality. And a draw is never a proof. –  Nov 06 '17 at 09:03
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Since $\cos(x)\leq 1$, one inequality follow. For the other one

Method 1

$$1-\cos(x)=\cos(0)-\cos(x)=2\sin\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)=2\sin^2\left(\frac{x}{2}\right)^2\leq \frac{x^2}{2},$$

where the last inequality come from the fact that $|\sin(x)|\leq |x|$ for all $x$.

Method 2

For all $x$, there is $c_x\in ]0,x[$ s.t. $$1-\cos(x)=\cos(c_x)\frac{x^2}{2}\leq \frac{x^2}{2}.$$

idm
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