# Proving this trigonometric inequality?

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

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

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\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
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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.

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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 $\alpha \in [0, \sqrt{2}]$.

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