Prove inequality $|e^{i\theta}-e^{i\phi}|\leq |\theta-\phi|$ for all $\theta,~\phi$ such that $0\leq \theta- \phi\leq \pi.$ 
Prove that $|e^{i\theta}-e^{i\phi}|\leq |\theta-\phi|$ for all $\theta,~\phi$ such that $0\leq \theta- \phi\leq \pi.$

Attempt. $|e^{i\theta}-e^{i\phi}|^2=(\cos\theta-\cos\phi)^2+(\sin\theta-\sin\phi)^2\leq 2(\theta-\phi)^2.$
How do we get rid of $2$ in the above inequality? 
Thanks in advance!
 A: This approach seems better.$$\begin{align} & |e^{i\theta}-e^{i\phi}|^2 \\ & =(\cos\theta-\cos\phi)^2+(\sin\theta-\sin\phi)^2 \\ & =2-2\cos(\theta-\phi) \\ & =2\left[1-\cos(\theta-\phi)\right] \\ & =2 \cdot 2 \sin^2\left(\frac{\theta-\phi}{2}\right)  \\ & \leq 4\left(\frac{\theta-\phi}{2}\right)^2 \text{since} \,\,\,\,\,\ \sin x \leq x  \,\,\,\,\, \forall \,\, x \in \left[0,\frac{\pi}{2}\right] \text{for } \,\,\, x=\frac{\theta-\phi}{2} \\ & \leq \left(\theta-\phi\right)^2 \end{align}$$
This implies $$\boxed{|e^{i\theta}-e^{i\phi}|^2\leq \left(\theta-\phi\right)^2 }$$
See if this helps.
A: How about "the line is the shortest path between two points in the usual metric of $\mathbb{C}$"?:
                         
To put it into numbers, parametrize the red curve by $\gamma(t)=(\cos t,\sin t),$ $t\in[\theta,\phi]$, and use the length-minimizing property of lines in the usual metric to get
$$
|e^{i\phi}-e^{i\theta}|\leq\text{length}(\gamma)=\int_\theta^\phi||\gamma'(t)||\,dt=\int_\theta^\phi\underbrace{||(-\sin t,\cos t)||}_{=1}\,dt=|\phi-\theta|.
$$
