Key Concepts and Formulas

• Complementary Angle Identities: $\sin(\theta) = \cos(\frac{\pi}{2} - \theta)$ and $\cos(\theta) = \sin(\frac{\pi}{2} - \theta)$.
• Half-Angle Identities: $1 + \cos(2\theta) = 2\cos^2(\theta)$ and $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$.
• Euler's Formula: $e^{i\theta} = \cos\theta + i\sin\theta$.
• De Moivre's Theorem: $(e^{i\theta})^n = e^{in\theta}$.

Step-by-Step Solution

Let the given expression be denoted by $E$: $E = {\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}$

Step 1: Transform trigonometric terms using complementary angles We want to express the numerator and denominator in terms of $\cos \theta + i \sin \theta$ to apply Euler's formula. We use the complementary angle identities to achieve this. Let $\theta = \frac{2\pi}{9}$. Then $\frac{\pi}{2} - \theta = \frac{\pi}{2} - \frac{2\pi}{9} = \frac{5\pi}{18}$. Therefore, $\sin \frac{2\pi}{9} = \cos \frac{5\pi}{18}$ and $\cos \frac{2\pi}{9} = \sin \frac{5\pi}{18}$.

Substituting these into the expression, we have: $E = {\left( {{{1 + \cos {{5\pi } \over {18}} + i\sin {{5\pi } \over {18}}} \over {1 + \cos {{5\pi } \over {18}} - i\sin {{5\pi } \over {18}}}}} \right)^3}$

Step 2: Apply half-angle identities Let $\alpha = \frac{5\pi}{18}$. We apply the half-angle identities to simplify $1 + \cos \alpha$ and $\sin \alpha$. We have $1 + \cos(2\theta) = 2\cos^2(\theta)$ and $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. Let $2\theta = \alpha = \frac{5\pi}{18}$, so $\theta = \frac{5\pi}{36}$. Then $1 + \cos \left( \frac{5\pi}{18} \right) = 2\cos^2 \left( \frac{5\pi}{36} \right)$ and $\sin \left( \frac{5\pi}{18} \right) = 2\sin \left( \frac{5\pi}{36} \right)\cos \left( \frac{5\pi}{36} \right)$.

Substituting these into the expression for $E$, we get: $E = {\left( {{{2\cos^2 \left( {{{5\pi } \over {36}}} \right) + i \left( {2\sin \left( {{{5\pi } \over {36}}} \right)\cos \left( {{{5\pi } \over {36}}} \right)} \right)} \over {2\cos^2 \left( {{{5\pi } \over {36}}} \right) - i \left( {2\sin \left( {{{5\pi } \over {36}}} \right)\cos \left( {{{5\pi } \over {36}}} \right)} \right)}}} \right)^3}$

Step 3: Factor and simplify Factor out $2\cos \left( \frac{5\pi}{36} \right)$ from both the numerator and denominator: $E = {\left( {{{2\cos \left( {{{5\pi } \over {36}}} \right) \left[ {\cos \left( {{{5\pi } \over {36}}} \right) + i\sin \left( {{{5\pi } \over {36}}} \right)} \right]} \over {2\cos \left( {{{5\pi } \over {36}}} \right) \left[ {\cos \left( {{{5\pi } \over {36}}} \right) - i\sin \left( {{{5\pi } \over {36}}} \right)} \right]}}} \right)^3}$ Since $\cos \left( \frac{5\pi}{36} \right) \neq 0$, we can cancel the common factor: $E = {\left( {{{\cos \left( {{{5\pi } \over {36}}} \right) + i\sin \left( {{{5\pi } \over {36}}} \right)} \over {\cos \left( {{{5\pi } \over {36}}} \right) - i\sin \left( {{{5\pi } \over {36}}} \right)}}} \right)^3}$

Step 4: Convert to Euler's form Let $\phi = \frac{5\pi}{36}$. Using Euler's formula, the numerator is $e^{i\phi}$ and the denominator is $e^{-i\phi}$. $E = {\left( {{{{e^{i{{5\pi } \over {36}}}}} \over {{e^{ - i{{5\pi } \over {36}}}}}}} \right)^3}$

Step 5: Simplify using exponent rules Using the rule $\frac{a^m}{a^n} = a^{m-n}$, we have: $E = {\left( {{e^{i{{5\pi } \over {36}} - \left( { - i{{5\pi } \over {36}}} \right)}}} \right)^3} = {\left( {{e^{i\left( {{{10\pi } \over {36}}} \right)}}} \right)^3} = {\left( {{e^{i{{5\pi } \over {18}}}}} \right)^3}$

Step 6: Apply De Moivre's Theorem Using De Moivre's Theorem, $(e^{i\theta})^n = e^{in\theta}$: $E = {e^{i\left( {{{5\pi } \over {18}} \times 3} \right)}} = {e^{i{{15\pi } \over {18}}}} = {e^{i{{5\pi } \over 6}}}$

Step 7: Convert back to rectangular form Using Euler's formula, $e^{i\theta} = \cos\theta + i\sin\theta$: $E = \cos \left( {{{5\pi } \over 6}} \right) + i\sin \left( {{{5\pi } \over 6}} \right)$ We know that $\cos \left( \frac{5\pi}{6} \right) = -\frac{\sqrt{3}}{2}$ and $\sin \left( \frac{5\pi}{6} \right) = \frac{1}{2}$. Therefore, $E = -\frac{\sqrt{3}}{2} + i\frac{1}{2} = \frac{1}{2}\left(-\sqrt{3} + i\right)$

Common Mistakes & Tips

• Double-check trigonometric identities, especially when dealing with half-angles.
• Ensure the complex number is in the $\cos\theta + i\sin\theta$ form before applying Euler's formula.
• Pay attention to the quadrant of the angle when evaluating trigonometric functions to determine the correct sign.

Summary

We simplified the given complex expression by using complementary angle identities, half-angle identities, Euler's formula, and De Moivre's Theorem. We transformed the expression into a simpler form, performed algebraic manipulations, and finally converted it back to rectangular form. The final value of the expression is $\frac{1}{2}(-\sqrt{3} + i)$.

Final Answer

The final answer is \boxed{\frac{1}{2}\left( {\sqrt 3 - i} \right)}, which corresponds to option (A).