Q48 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics

$\operatorname{Arg}\left(\sin \frac{6 \pi}{5}+i\left(1+\cos \frac{6 \pi}{5}\right)\right)=$

  • A
    $\frac{5 \pi}{6}$
  • B
    $\frac{6 \pi}{5}$
  • C
    $\frac{2 \pi}{5}$
  • Correct answer
    D
    $\frac{9 \pi}{10}$

Solution

$\begin{aligned} & \operatorname{Arg}\left(\sin \frac{6 \pi}{5}+i\left(1+\cos \frac{6 \pi}{5}\right)\right) \\ = & \operatorname{Arg}\left(2 \sin \frac{3 \pi}{5} \cdot \cos \frac{3 \pi}{5}+i \cdot 2 \cos ^2 \frac{3 \pi}{5}\right) \\ = & \operatorname{Arg}\left[\left(2 \cos \frac{3 \pi}{5}\right)\left(\sin \frac{3 \pi}{5}+i \cos \frac{3 \pi}{5}\right)\right] \\ = & \operatorname{Arg}\left[\left(2 \cos \frac{3 \pi}{5}\right)\left(\sin \left(\frac{\pi}{2}+\frac{\pi}{10}\right)+i \cos \left(\frac{\pi}{2}+\frac{\pi}{10}\right)\right)\right] \\ = & \operatorname{Arg}\left[\left(2 \cos \frac{3 \pi}{5}\right)\left(\cos \frac{\pi}{10}-i \sin \frac{\pi}{10}\right)\right] \\ \therefore & \quad \operatorname{Arg}\left(Z_1 \cdot Z_2\right)=\arg Z_1+\arg Z_2\end{aligned}$ $=0-\frac{\pi}{10}=\frac{-\pi}{10}$ or $\pi-\frac{\pi}{10}=\frac{9 \pi}{10}$.

Hence, the correct answer is option: 'D'.

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