Complex number identity by trigonometry Show that $\lvert e^{i\theta} - 1\rvert = 2\lvert\sin(\theta/2)\rvert$ by using the geometry of the triangle with vertices 0, 1, and the midpoint of the line joining 0 and $e^{i\theta}$.
I have been able to show this identity through other means, however I am stuck on how to utilize this particular triangle. 

$|e^{i\theta}-1|=|\cos\theta+i\sin\theta-(1+0i)|=|(\cos\theta-1)+i\sin\theta|=\sqrt{(\cos\theta-1)^2+(\sin\theta)^2}=\sqrt{\cos^2\theta+1-2\cos\theta+\sin^2\theta}=\sqrt{2-2\cos\theta}=\sqrt{2(1-\cos\theta)}=\sqrt{2(\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2}-\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2})}=\sqrt{4sin^2\frac{\theta}{2}}=2|\sin\frac{\theta}{2}|$
