The argument of the complex number sin ,frac{ | vedclass

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Question

$\sin \frac{6 \pi}{5}+i\left(1+\cos \frac{6 \pi}{5}\right)$

lies in ${2^{nd}}$ quadrant and

$\left|\frac{1+\cos \frac{6 \pi}{5}}{\sin \frac{6 \p

Vedclass · Accepted Answer

$\frac{{9\pi }}{10}$

Vedclass · Answer

$\sin \frac{6 \pi}{5}+i\left(1+\cos \frac{6 \pi}{5}ight)$

lies in ${2^{nd}}$ quadrant and

$\left|\frac{1+\cos \frac{6 \pi}{5}}{\sin \frac{6 \pi}{5}}ight|=\left|\cot \left(\frac{3 \pi}{5}ight)ight|=\left|\cot \left(\frac{\pi}{2}+\frac{\pi}{10}ight)ight|=	an \frac{\pi}{10}$

${2^{nd}}$ quadrant $\Rightarrow \pi-\frac{\pi}{10}$

The argument of the complex number $\sin \,\frac{{6\pi }}{5}\, + \,i\,\left( {1\, + \,\cos \,\frac{{6\pi }}{5}} \right)$ is

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