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This question already has an answer here:

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?

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ $\left|\sin(\dfrac t 2)\right|=\left|\dfrac 12(e^{i\frac t2}\color{red}-e^{-i\frac t2)})\right|$ $\endgroup$ – J. W. Tanner Jun 23 at 11:38
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$\begin{align} |e^{it}-1|^2&=(e^{it}-1)(\overline{e^{it}-1})\\ &=(e^{it}-1)(e^{-it}-1)\\ &=2-(e^{it}+e^{-it})\\ &=2-2 \cos t\\ &=4\sin^2\left(\frac{t}2\right) \end{align}$

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

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

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  • $\begingroup$ using $|zw|=|z||w|$ and $|e^{ix}|=|i|=1$ $\endgroup$ – J. W. Tanner Jun 23 at 11:45
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$|\cos(t)-1+i\sin(t)|=\sqrt{(\cos(t)-1)^2+(\sin(t))^2}=\sqrt{\cos^2(t)-2\cos(t)+1+\sin^2(t)}=\sqrt{2-2\cos(t)}=\sqrt{2(2\sin^2(\frac{t}{2})}=\sqrt{4\sin^2(\frac{t}{2})}=|2\sin(\frac{t}{2})|$

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    $\begingroup$ Put "\" before sin, cos, tan, log and etc. to make them appear neater. $\endgroup$ – DonAntonio Jun 23 at 11:37
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$$ \newcommand \abs [1]{\left\vert #1 \right\vert} \newcommand \imu {\mathrm i} \newcommand \rme {\mathrm e} \abs {\rme^{\imu t} - 1 } = \abs {\rme^{\imu t/2}} \abs {\rme^{\imu t/2} - \rme^{-\imu t/2}} = 2 \abs {\frac {\rme^{\imu t/2} - \rme^{-\imu t/2}} {2\imu}} = 2 \abs { \sin (t/2)}\quad \mathrm {Q.E.D.} $$

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

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