# Use of inequality $1 - \cos (x) \leq x^2 /2$

In this answer the value of $1 - \cos(x)$ has to be evaluated in order to find its upper limit, if it exists.

In particular, $x = 2 \pi / n$. The answer is related to the length of a side of a regular $n$-gon inscribed into a unit-radius circumference; because the perimeter of the $n$-gon is always less than $2 \pi$, the single side must always be less than $2 \pi / n$.

The inequality

$$1 - \cos (x) \leq \displaystyle \frac{x^2}{2}$$ (1)

is used and the proof is completed with

$$2(1 - \cos(x)) \leq (2 \pi / n)^2$$

$$\sqrt{2(1 - \cos(x))} \leq 2 \pi / n$$

But it is well known that the cosine is a function $f(x) \in [-1;1]$, so $1 - \cos (x) \in [0,2]$. By using this information, we would obtain

$$1 - \cos (x) \leq 2$$ (2)

The proof would provide

$$2(1 - \cos(x)) \leq 4$$

$$\sqrt{2(1 - \cos(x))} \leq 2$$

which is a completely different result.

• Why in that case it is preferable to use (1) instead of (2)?
• How to choose when it is convenient to use (1) and when to use (2) in a proof?
• Strange question. The inequality $(2)$ is trivial and doesn't help addressing the $n$-gon problem.
– user65203
Commented Dec 9, 2017 at 16:16

The bound $$-1\le \cos(x)\le 1$$ is correct, but rather a crude one.

We can easily obtain tighter bounds using the inequality from elementary geometry

$$|\sin(\theta)|\le |\theta| \tag 1$$

for all $$\theta$$, and the half-angle formula for the sine function

$$1-\cos(\theta)=2\sin^2(\theta/2) \tag 2$$

Squaring both sides of $$1$$, substituting $$\theta =x/2$$, and using $$(2)$$ reveals

\begin{align} \sin^2(x/2)&=\frac{1-\cos(x)}{2}\\\\ &\le x^2/4 \tag 3 \end{align}

whereupon we find for all $$x$$

$$\bbox[5px,border:2px solid #C0A000]{1-\cos(x)\le \frac12 x^2} \tag 4$$

For values of $$x<2$$, $$\frac12 x^2<2$$ and $$(4)$$ provides a tighter bound than $$1-\cos(x)\le 2$$. For $$x\ge 2$$, it is still correct that $$1-\cos(x)\le \frac12 x^2$$, but the inequality $$1-\cos(x)\le 2$$ is obviously tighter. Therefore, we can write

$$1-\cos(x)\le \begin{cases}\frac12 x^2&,x<2\\\\2&,x\ge 2\tag 5\end{cases}$$

So, $$(5)$$ provides a guideline for the appropriate use of the bounds for $$1-\cos(x)$$.

• Yesterday I open this question and when I realize result was irrelevant, I did not even read the whole. Clearly, I made a mistake and I am struggling to understand how. But in any case I would like to see 'Would the downwoter care to comment?'. Cowardly? Please be respectful. Commented Dec 9, 2017 at 2:24
• Unfortunately, I am unable to redo downvote. Please make an unnoticeable edit if possible. Commented Dec 9, 2017 at 2:24
• @atbey I've edited and removed the comment. Commented Dec 9, 2017 at 16:13
• @atbey Much appreciative! Commented Dec 9, 2017 at 20:02

Let $x\gt 0$. You know that $\sin(x) < x$ for such $x$. Integrating $$1 - \cos(x) =\int_0^x \sin(t)\, dt < \int_0^x t\,dt = {t^2\over 2}$$

ncmathsadist's technique can be extended for a tighter bound as described below. It also extends the region over which the bound is better than $$1 - \cos (x) \leq 2$$ per Mark Viola's answer.

Cutting off the Taylor series for $$\sin(t)$$ before a negative term (the terms alternate sign), one gets a polynomial bound on $$\sin(t)$$ that can be integrated.

For example $$\sin (t) \leq t-t^3/3!+t^5/5!$$, therefore

$$1-\cos (x)=\int_0^x \sin (t)\ dt \leq \int_0^x t-\frac{t^3}{3!}+\frac{t^5}{5!}\ dt = \frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!}$$

The cut-off approximation is an upper bound. The cut-off residual will always be negative. This can be shown by summing the terms pairwise. Since $$\sin (t)$$ is absolutely convergent, there isn't an issue with ordering of terms.

A bound which is better than $$1 - \cos (x) \leq 2$$ from $$[0,2\,\pi]$$ is: $$1-\cos (x) \leq \frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} - \frac{x^{8}}{8!} + \frac{x^{10}}{10!} - \frac{x^{12}}{12!} + \frac{x^{14}}{14!} \leq 2$$

You can then see that all this really says is: $$\cos (x) \geq 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^{8}}{8!} - \frac{x^{10}}{10!} + \frac{x^{12}}{12!} - \frac{x^{14}}{14!}$$ ...or $$\cos (x)$$ is greater than its Taylor expansion with the cut-off starting at a positive term.