Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $|\theta-\theta_0|\leqslant \frac{\pi}4$.

How can I prove that $$2(1-\cos(\theta-\theta_0))\geqslant \frac{|\theta-\theta_0|^2}{2}?$$

share|cite|improve this question
\begin{align} 2(1-\cos(\theta-\theta_0))=\sin^2\left(\dfrac{\theta-\theta_0}{2}\right)\\ \end{align} – Inquest Dec 5 '12 at 16:50
The fact mentioned by Inquest may be more familiar to you as $\cos 2\alpha=2\cos^2\alpha -1=1-2\sin^2\alpha$. From $\cos 2\alpha=1-2\sin^2\alpha$, by rearranging, you will get desired expression for $1-\cos(\theta-\theta_0)$. – André Nicolas Dec 5 '12 at 16:56
oh yes, ok, i simply use taylor expansion, thank you very much – bateman Dec 5 '12 at 16:57
@Inquest: Replace 2 in LHS by 1/2. – Gautam Shenoy Dec 5 '12 at 17:18
@Inquest The correct identity is $$1-\cos \left( \theta -\theta _{0}\right) =2\sin ^{2}\frac{\theta -\theta _{0}}{2}$$ – Américo Tavares Dec 5 '12 at 18:19
up vote 0 down vote accepted

Let $\alpha=\dfrac{|\theta-\theta_0|}{2}$.

By a trigonometric identity discussed in comments, we need to show that $\sin \alpha \ge \frac{\alpha}{\sqrt{2}}$.

This is true with room to spare. Let $f(x)=\sin x-\dfrac{x}{\sqrt{2}}$.

We have $f(0)=0$. And $f'(x)=\cos x-\dfrac{1}{\sqrt{2}}$. Note that $f'(x)$ is positive until $x=\pi/4$. So $f(x)$ is increasing in $[0,\pi/4]$, and is therefore $\ge 0$ in this interval, and somewhat beyond.

share|cite|improve this answer

Let $\alpha \in [0, \frac{\pi}{2}]$ (so a larger interval than requested). Draw an arc of angle $\alpha$ on the unit circle, starting at $(1,0)$. The length of the chord between the endpoints squared is $$\ell^2 = \sin(\alpha)^2 + (1-\cos(\alpha))^2 = 2 - 2\cos(\alpha).$$ Since the length of the chord is at most the length of the arc you get the inequality

$$2 - 2\cos(\alpha) \leq \alpha^2$$

which is interesting but the wrong way around. Now let $d$ be the distance from $(0,0)$ to the (centre of the) chord and draw an arc of angle $\alpha$ but with a smaller radius $d$. Then this smaller arc touches the chord from the inside and has a length that is at most the length of the chord. This shows that

$$ 2 - 2\cos(\alpha) \geq d^2 \alpha^2 $$

and since $d$ is at least $\frac{1}{\sqrt{2}}$ your inequality follows. In fact $$d^2 = 1 - \frac{\ell^2}{4} = \frac{1 + \cos(\alpha)}{2}$$ and together with the inequalities so far we get

$$ 2 - 2\cos(\alpha) \geq \frac{1 + \cos(\alpha)}{2} \alpha^2 \geq \frac{1 + 1 - \frac{\alpha^2}{2}}{2} \alpha^2 = \alpha^2 - \frac{\alpha^4}{4}. $$

which is a sharper result for $\alpha \in [0, \sqrt{2}]$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.