# How to prove $\big | e^{it}-1 \big |=2 \bigg |\sin \left(\frac{t}{2}\right) \bigg |$ [duplicate]

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?

## marked as duplicate by Martin R, YuiTo Cheng, Somos, Lee David Chung Lin, cmkJun 23 at 15:00

• $\left|\sin(\dfrac t 2)\right|=\left|\dfrac 12(e^{i\frac t2}\color{red}-e^{-i\frac t2)})\right|$ – J. W. Tanner Jun 23 at 11:38

\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}

• Neat, precise, elegant and beautiful. +1 – DonAntonio Jun 23 at 11:38
• Thank you @DonAntonio!! – Thomas Shelby Jun 23 at 11:39
• In Thomas' answer it is $4\sin^2\left(\frac{t}2\right)$ – ParabolicAlcoholic Jun 23 at 13:41
• Ok, I understand. – ParabolicAlcoholic Jun 23 at 18:22

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 $$\sin$$?

$$|e^{it}-1|=|e^{it/2}||e^{it/2}-e^{-it/2}|=|2i|\left|\sin\left(\frac t2\right)\right|=2\left|\sin\left(\frac t2\right)\right|$$

• using $|zw|=|z||w|$ and $|e^{ix}|=|i|=1$ – J. W. Tanner Jun 23 at 11:45

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

• Put "\" before sin, cos, tan, log and etc. to make them appear neater. – DonAntonio Jun 23 at 11:37

$$\newcommand \abs [1]{\left\vert #1 \right\vert} \newcommand \imu {\mathrm i} \newcommand \rme {\mathrm e} \abs {\rme^{\imu t} - 1 } = \abs {\rme^{\imu t/2}} \abs {\rme^{\imu t/2} - \rme^{-\imu t/2}} = 2 \abs {\frac {\rme^{\imu t/2} - \rme^{-\imu t/2}} {2\imu}} = 2 \abs { \sin (t/2)}\quad \mathrm {Q.E.D.}$$

$$\mid z\mid$$ 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 $$2|\sin(t/2)|$$.