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