The argument of the complex number sin frac{6 | vedclass

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(C) Let $z = \sin \frac{6\pi}{5} + i(1 + \cos \frac{6\pi}{5})$.Using the identities $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ and

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

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(C) Let $z = \sin \frac{6\pi}{5} + i(1 + \cos \frac{6\pi}{5})$.Using the identities $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ and $1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$,we have:$z = 2 \sin \frac{3\pi}{5} \cos \frac{3\pi}{5} + i(2 \cos^2 \frac{3\pi}{5})$$z = 2 \cos \frac{3\pi}{5} (\sin \frac{3\pi}{5} + i \cos \frac{3\pi}{5})$Since $\cos \frac{3\pi}{5}$ is negative,we write $z = -2 \cos \frac{3\pi}{5} (-\sin \frac{3\pi}{5} - i \cos \frac{3\pi}{5})$.Using $-\sin 	heta = \cos(\frac{\pi}{2} + 	heta)$ and $-\cos 	heta = \sin(\frac{\pi}{2} + 	heta)$,we get:$z = |z| (\cos(\frac{\pi}{2} + \frac{3\pi}{5}) + i \sin(\frac{\pi}{2} + \frac{3\pi}{5}))$$z = |z| (\cos \frac{11\pi}{10} + i \sin \frac{11\pi}{10})$.Alternatively,the argument $	heta$ satisfies $	an 	heta = \frac{1 + \cos(6\pi/5)}{\sin(6\pi/5)} = \frac{2 \cos^2(3\pi/5)}{2 \sin(3\pi/5) \cos(3\pi/5)} = \cot(3\pi/5) = 	an(\frac{\pi}{2} - \frac{3\pi}{5}) = 	an(-\frac{\pi}{10})$.Since the real part is negative and the imaginary part is negative,the complex number lies in the $3^{rd}$ quadrant.Thus,$	heta = \pi - \frac{\pi}{10} = \frac{9\pi}{10}$.

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

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