First, we are given the expression: \(\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^3\). To solve this, let's rewrite the expression inside the parentheses using the properties of complex numbers and simplification techniques.
Consider the expression: \(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\). By multiplying the numerator and the denominator by the conjugate of the denominator, we simplify the expression as follows:
Multiply numerator and denominator by \(1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}\) (conjugate of the denominator): \(\frac{(1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9})^2}{(1+\sin \frac{2 \pi}{9})^2+\cos^2 \frac{2 \pi}{9}}\).
Simplify the denominator using the identity \(\cos^2 x + \sin^2 x = 1\), which gives: \((1+\sin \frac{2 \pi}{9})^2+\cos^2 \frac{2 \pi}{9} = 2+2\sin \frac{2 \pi}{9}\).
The expression now looks like this: \(\frac{(1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9})^2}{2+2\sin \frac{2 \pi}{9}}\).
Recall that \(\sin^2 x - \cos^2 x = -\cos(2x)\), thus the real part becomes \(1 + 2\sin \frac{2 \pi}{9} - \cos \left(\frac{4\pi}{9}\right)\). The imaginary part becomes \(2i\cos \frac{2 \pi}{9}(1+\sin \frac{2 \pi}{9})\).
Simplifying further using trigonometry identities and properties, we conclude: \(\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^3 = -\frac{1}{2}(\sqrt{3}-i)\).
Thus, the correct option is \(-\frac{1}{2}(\sqrt{3}-i)\).
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Top Questions on Some Properties of Definite Integrals
