Proving this trigonometric inequality? Let $|\theta-\theta_0|\leqslant \frac{\pi}4$.
How can I prove that $$2(1-\cos(\theta-\theta_0))\geqslant \frac{|\theta-\theta_0|^2}{2}?$$
 A: 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 $\alpha \in [0, \sqrt{2}]$.
A: 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. 
