How to prove $\big | e^{it}-1 \big |=2 \bigg |\sin \left(\frac{t}{2}\right) \bigg |$ i started by rewriting:$$\big | \cos(t)+i\sin(t)-1 \big |=2 \bigg |\sin \left(\frac{t}{2}\right) \bigg |$$$$\bigg | \frac{1}{2}(e^{it}+e^{-it})+\frac{1}2(e^{it}-e^{-it})-1 \bigg |=2 \bigg |\frac{1}{2}(e^{i\frac{t}{2}}+e^{-i\frac{t}{2}}) \bigg |$$ How do I get rid of the absolute value bars?
 A: $\begin{align}
|e^{it}-1|^2&=(e^{it}-1)(\overline{e^{it}-1})\\
&=(e^{it}-1)(e^{-it}-1)\\
&=2-(e^{it}+e^{-it})\\
&=2-2 \cos t\\
&=4\sin^2\left(\frac{t}2\right)
\end{align}$
A: Use that $$
e^{i\theta}-1=e^{i\theta/2}(e^{i\theta/2}-e^{-i\theta/2})
$$
Then what is $|e^{i\theta/2}|$ and how can we express $e^{i\theta/2}-e^{-i\theta/2}$ as a $\sin$? 
A: $|\cos(t)-1+i\sin(t)|=\sqrt{(\cos(t)-1)^2+(\sin(t))^2}=\sqrt{\cos^2(t)-2\cos(t)+1+\sin^2(t)}=\sqrt{2-2\cos(t)}=\sqrt{2(2\sin^2(\frac{t}{2})}=\sqrt{4\sin^2(\frac{t}{2})}=|2\sin(\frac{t}{2})|$
A: $$|e^{it}-1|=|e^{it/2}||e^{it/2}-e^{-it/2}|=|2i|\left|\sin\left(\frac t2\right)\right|=2\left|\sin\left(\frac t2\right)\right|$$
A: $$
\newcommand \abs [1]{\left\vert #1 \right\vert}
\newcommand \imu {\mathrm i}
\newcommand \rme {\mathrm e}
\abs {\rme^{\imu t} - 1 } = \abs {\rme^{\imu t/2}} \abs {\rme^{\imu t/2} - \rme^{-\imu t/2}} = 2 \abs {\frac {\rme^{\imu t/2} - \rme^{-\imu t/2}}  {2\imu}} = 2 \abs {
\sin (t/2)}\quad \mathrm {Q.E.D.} 
$$
A: $\mid z\mid$ is the modulus of the complex number $z=x+iy$ defined to be
$$\mid z\mid=\mid x+iy\mid=\sqrt{x^2+y^2},$$
essentially the length of $z$.
So in your case, expanding $|\cos t-1+i\sin t|^2$, we have
$$(\cos t-1)^2+\sin^2t=\cos^2t-2\cos t+1+\sin^2t=2(1-\cos t)=4\sin^2(t/2).$$
Taking square roots, you get $2|\sin(t/2)|$.
