Question

The value of ${\left(\frac{1+\sin {\displaystyle \frac{2\mathrm{\pi }}{9}}+i\cos {\displaystyle \frac{2\mathrm{\pi }}{9}}}{1+\sin {\displaystyle \frac{2\mathrm{\pi }}{9}}-i\cos {\displaystyle \frac{2\mathrm{\pi }}{9}}}\right)}^3$ is

• A. $\frac{-1}{2}\left(1-i\sqrt{3}\right)$
• B. $\frac{1}{2}\left(1-i\sqrt{3}\right)$
• C. $\frac{-1}{2}\left(\sqrt{3}-i\right)$
• D. $\frac{1}{2}\left(\sqrt{3}-i\right)$

Answer 1

The correct option is C

$\frac{-1}{2}\left(\sqrt{3}-i\right)$

Explanation for the correct answer:

Find the value of the given trigonometric expression

Let $S={\left(\frac{1+\sin {\displaystyle \frac{2\mathrm{\pi }}{9}}+i\cos {\displaystyle \frac{2\mathrm{\pi }}{9}}}{1+\sin {\displaystyle \frac{2\mathrm{\pi }}{9}}-i\cos {\displaystyle \frac{2\mathrm{\pi }}{9}}}\right)}^3$

$\Rightarrow$ $S={\left(\frac{1+\cos {\displaystyle \left(\frac{\mathrm{\pi }}{2}-\frac{2\mathrm{\pi }}{9}\right)}+i\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-{\displaystyle \frac{2\mathrm{\pi }}{9}}\right)}{1+\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-{\displaystyle \frac{2\mathrm{\pi }}{9}}\right)-i\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-{\displaystyle \frac{2\mathrm{\pi }}{9}}\right)}\right)}^3$ ; $$\begin{gathered} \left[\because sin\left(\frac{\pi }{2}-x\right)=cosx\right] \\ \left[\because cos\left(\frac{\pi }{2}-x\right)=sinx\right] \end{gathered}$$

$\Rightarrow$ $S={\left(\frac{1+\cos {\displaystyle \left(\frac{5\mathrm{\pi }}{18}\right)}+i\sin \left({\displaystyle \frac{5\mathrm{\pi }}{18}}\right)}{1+\cos \left({\displaystyle \frac{5\mathrm{\pi }}{18}}\right)-i\sin \left({\displaystyle \frac{5\mathrm{\pi }}{18}}\right)}\right)}^3$

$\Rightarrow$ $S={\left(\frac{\cancel{1}+2{\cos }^2\left({\displaystyle \frac{5\mathrm{\pi }}{36}}\right)-\cancel{1}+2i\sin \left({\displaystyle \frac{5\mathrm{\pi }}{36}}\right)\cos \left({\displaystyle \frac{5\mathrm{\pi }}{36}}\right)}{\cancel{1}+2{\cos }^2\left({\displaystyle \frac{5\mathrm{\pi }}{36}}\right)-\cancel{1}-2i\sin \left({\displaystyle \frac{5\mathrm{\pi }}{36}}\right)\cos \left({\displaystyle \frac{5\mathrm{\pi }}{36}}\right)}\right)}^3$ ; $$\begin{gathered} \left[\because cos2x=2co{s}^{2}x-1\right] \\ \left[\because sin2x=2sinxcosx\right] \end{gathered}$$

$\Rightarrow$ $S={\left(\frac{\cancel{2\cos \left({\displaystyle \frac{5\mathrm{\pi }}{36}}\right)}\left(\cos \left({\displaystyle \frac{5\mathrm{\pi }}{36}}\right)+i\sin \left({\displaystyle \frac{5\mathrm{\pi }}{36}}\right)\right)}{\cancel{2\cos \left({\displaystyle \frac{5\mathrm{\pi }}{36}}\right)}\left(\cos \left({\displaystyle \frac{5\mathrm{\pi }}{36}}\right)-i\sin \left({\displaystyle \frac{5\mathrm{\pi }}{36}}\right)\right)}\right)}^3$

$\Rightarrow$ $S={\left(\frac{e^{{\displaystyle \frac{5i\mathrm{\pi }}{36}}}}{e^{{\displaystyle \frac{-5i\mathrm{\pi }}{36}}}}\right)}^3$ ; $\left[\because e^{i\theta }=cos\theta +isin\theta \right]$

$\Rightarrow$ $S={\left(e^{\frac{5i\mathrm{\pi }}{18}}\right)}^3$

$\Rightarrow$ $S=e^{i\frac{5\mathrm{\pi }}{6}}$

Replace, ${\mathrm{e}}^{\mathrm{i}\frac{\mathrm{5}\mathrm{\pi }}{\mathrm{6}}}\mathrm{=}\cos \frac{\mathrm{5}\mathrm{\pi }}{\mathrm{6}}\mathrm{+}\mathrm{isin}\frac{\mathrm{5}\mathrm{\pi }}{\mathrm{6}}$, we get,

$\Rightarrow$ $S=\cos \frac{5\mathrm{\pi }}{6}+i\sin \frac{5\mathrm{\pi }}{6}$

$\Rightarrow$ $S=\frac{-\sqrt{3}}{2}+\frac{i}{2}$

$\Rightarrow$ $S=\frac{-1}{2}\left(\sqrt{3}-i\right)$.

Hence, option $\left(C\right)$, $\frac{-1}{2}\left(\sqrt{3}-i\right)$ is the correct answer.