If `a=cos""(4pi)/(3)+isin""(4pi)/(3)`, then the value of `((1+a)/(2))^(3n)` is :

Options

A. `(-1)^(n)`
B. `(-1)^(n)/(2^(3n))`
C. `(1)/(2^(3n))`
D. `(-1)^(n)+1`

Answer

To solve the problem, we start with the expression given for \( a \):

\[ a = \cos\left(\frac{4\pi}{3}\right) + i \sin\left(\frac{4\pi}{3}\right) \]

### Step 1: Rewrite \( a \) using the properties of trigonometric functions

We can express \( a \) in terms of \( \cos \) and \( \sin \) of \( \frac{4\pi}{3} \):

\[ \cos\left(\frac{4\pi}{3}\right) = \cos\left(\pi + \frac{\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2} \]

\[ \sin\left(\frac{4\pi}{3}\right) = \sin\left(\pi + \frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \]

Thus, we have:

\[ a = -\frac{1}{2} - i \frac{\sqrt{3}}{2} \]

### Step 2: Calculate \( 1 + a \)

Now, we need to calculate \( 1 + a \):

\[ 1 + a = 1 - \frac{1}{2} - i \frac{\sqrt{3}}{2} = \frac{1}{2} - i \frac{\sqrt{3}}{2} \]

### Step 3: Divide by 2

Next, we find \( \frac{1 + a}{2} \):

\[ \frac{1 + a}{2} = \frac{\frac{1}{2} - i \frac{\sqrt{3}}{2}}{2} = \frac{1}{4} - i \frac{\sqrt{3}}{4} \]

### Step 4: Express in polar form

To simplify our calculations, we express \( \frac{1 + a}{2} \) in polar form. The modulus \( r \) is given by:

\[ r = \sqrt{\left(\frac{1}{4}\right)^2 + \left(-\frac{\sqrt{3}}{4}\right)^2} = \sqrt{\frac{1}{16} + \frac{3}{16}} = \sqrt{\frac{4}{16}} = \frac{1}{2} \]

The argument \( \theta \) is:

\[ \theta = \tan^{-1}\left(\frac{-\frac{\sqrt{3}}{4}}{\frac{1}{4}}\right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \]

Thus, we can write:

\[ \frac{1 + a}{2} = \frac{1}{2} \left( \cos\left(-\frac{\pi}{3}\right) + i \sin\left(-\frac{\pi}{3}\right) \right) \]

### Step 5: Raise to the power \( 3n \)

Now we raise \( \frac{1 + a}{2} \) to the power \( 3n \):

\[ \left(\frac{1 + a}{2}\right)^{3n} = \left(\frac{1}{2}\right)^{3n} \left( \cos\left(-\frac{3n\pi}{3}\right) + i \sin\left(-\frac{3n\pi}{3}\right) \right) \]

This simplifies to:

\[ \left(\frac{1}{2}\right)^{3n} \left( \cos(-n\pi) + i \sin(-n\pi) \right) \]

Using the periodic properties of cosine and sine:

\[ \cos(-n\pi) = (-1)^n, \quad \sin(-n\pi) = 0 \]

Thus, we have:

\[ \left(\frac{1 + a}{2}\right)^{3n} = \left(\frac{1}{2}\right)^{3n} (-1)^n \]

### Final Result

The final result is:

\[ \left(\frac{1 + a}{2}\right)^{3n} = \frac{(-1)^n}{2^{3n}} \]