Key Concepts and Formulas

• Euler's Formula: $e^{i\theta} = \cos\theta + i\sin\theta$
• De Moivre's Theorem: $(r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta))$, or equivalently, $(re^{i\theta})^n = r^n e^{in\theta}$
• Trigonometric Identities:
  • $\sin x = \cos\left(\frac{\pi}{2} - x\right)$
  • $\cos x = \sin\left(\frac{\pi}{2} - x\right)$
  • $1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)$
  • $\sin\theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$

Step-by-Step Solution

Step 1: Rewrite using co-function identities We begin by using the co-function identities to express the sine and cosine terms in a more suitable form for applying half-angle formulas.

Let $x = \frac{2\pi}{9}$. Then $\frac{\pi}{2} - x = \frac{\pi}{2} - \frac{2\pi}{9} = \frac{9\pi - 4\pi}{18} = \frac{5\pi}{18}$. So, $\sin\left(\frac{2\pi}{9}\right) = \cos\left(\frac{5\pi}{18}\right)$ and $\cos\left(\frac{2\pi}{9}\right) = \sin\left(\frac{5\pi}{18}\right)$. We substitute these into the given expression: $Z = {\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3} = {\left( {{{1 + \cos {{5\pi } \over 18} + i\sin {{5\pi } \over 18}} \over {1 + \cos {{5\pi } \over 18} - i\sin {{5\pi } \over 18}}}} \right)^3}$ Explanation: This step allows us to use the half-angle identities in the next step.

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

Numerator: $1 + \cos\left(\frac{5\pi}{18}\right) + i\sin\left(\frac{5\pi}{18}\right) = 2\cos^2\left(\frac{5\pi}{36}\right) + i\left(2\sin\left(\frac{5\pi}{36}\right)\cos\left(\frac{5\pi}{36}\right)\right) = 2\cos\left(\frac{5\pi}{36}\right)\left(\cos\left(\frac{5\pi}{36}\right) + i\sin\left(\frac{5\pi}{36}\right)\right)$ Denominator: $1 + \cos\left(\frac{5\pi}{18}\right) - i\sin\left(\frac{5\pi}{18}\right) = 2\cos^2\left(\frac{5\pi}{36}\right) - i\left(2\sin\left(\frac{5\pi}{36}\right)\cos\left(\frac{5\pi}{36}\right)\right) = 2\cos\left(\frac{5\pi}{36}\right)\left(\cos\left(\frac{5\pi}{36}\right) - i\sin\left(\frac{5\pi}{36}\right)\right)$ Explanation: Using the half-angle identities helps us to factor out a common term and express the numerator and denominator in terms of cosine and sine of the same angle.

Step 3: Convert to Euler form Using Euler's formula, we can rewrite the trigonometric expressions in exponential form. Numerator: $2\cos\left(\frac{5\pi}{36}\right)\left(\cos\left(\frac{5\pi}{36}\right) + i\sin\left(\frac{5\pi}{36}\right)\right) = 2\cos\left(\frac{5\pi}{36}\right)e^{i\frac{5\pi}{36}}$ Denominator: $2\cos\left(\frac{5\pi}{36}\right)\left(\cos\left(\frac{5\pi}{36}\right) - i\sin\left(\frac{5\pi}{36}\right)\right) = 2\cos\left(\frac{5\pi}{36}\right)e^{-i\frac{5\pi}{36}}$ Explanation: Applying Euler's formula simplifies the expression and makes it easier to manipulate.

Step 4: Simplify the fraction Substitute the Euler forms back into the expression for $Z$: $Z = {\left( \frac{2\cos\left(\frac{5\pi}{36}\right) e^{i \frac{5\pi}{36}}}{2\cos\left(\frac{5\pi}{36}\right) e^{-i \frac{5\pi}{36}}} \right)^3} = {\left( \frac{e^{i \frac{5\pi}{36}}}{e^{-i \frac{5\pi}{36}}} \right)^3} = {\left( e^{i \frac{5\pi}{36} - \left(-i \frac{5\pi}{36}\right)} \right)^3} = {\left( e^{i \frac{10\pi}{36}} \right)^3} = {\left( e^{i \frac{5\pi}{18}} \right)^3}$ Explanation: Dividing complex numbers in exponential form involves subtracting the arguments.

Step 5: Apply De Moivre's Theorem Using De Moivre's Theorem, we raise the complex number to the power of 3: $Z = \left( e^{i \frac{5\pi}{18}} \right)^3 = e^{i \left(3 \times \frac{5\pi}{18}\right)} = e^{i \frac{15\pi}{18}} = e^{i \frac{5\pi}{6}}$ Explanation: De Moivre's Theorem simplifies raising a complex number in exponential form to a power.

Step 6: Convert back to rectangular form Convert $e^{i \frac{5\pi}{6}}$ back to the rectangular form using Euler's formula: $Z = \cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} + i\frac{1}{2} = -\frac{1}{2}\left(\sqrt{3} - i\right)$ Explanation: We convert back to rectangular form to match the options given.

Common Mistakes & Tips:

• Sign Errors: Be careful with the signs of sine and cosine in different quadrants.
• Trigonometric Values: Remember the standard trigonometric values for angles like $\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}$.
• Complex Conjugates: Recognize when the denominator is the complex conjugate of the numerator to simplify the expression.

Summary: We simplified the given expression by first using co-function identities and half-angle identities to rewrite the numerator and denominator in a form suitable for Euler's formula. Then, we converted to exponential form, simplified the fraction, applied De Moivre's Theorem, and finally converted back to rectangular form to obtain the final answer.

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