For the equation `(1-ix)/(1+ix)=sin.(pi)/(7)-i cos.(pi)/(7)`, if `x=tan((kpi)/(28))`, then the value of k can be (where `i^(2)=-1`)

Options

A. 1
B. 3
C. 5
D. 9

Answer

To solve the equation

\[ \frac{1 - ix}{1 + ix} = \sin\left(\frac{\pi}{7}\right) - i \cos\left(\frac{\pi}{7}\right), \]

where \( x = \tan\left(\frac{k\pi}{28}\right) \), we will follow these steps:

### Step 1: Rationalize the left-hand side

We start with the left-hand side:

\[ Z = \frac{1 - ix}{1 + ix}. \]

To rationalize, we multiply the numerator and denominator by the conjugate of the denominator:

\[ Z = \frac{(1 - ix)(1 - ix)}{(1 + ix)(1 - ix)} = \frac{(1 - ix)^2}{1 + x^2}. \]

### Step 2: Expand the numerator

Now we expand the numerator:

\[ (1 - ix)^2 = 1 - 2ix + (ix)^2 = 1 - 2ix - x^2. \]

Thus, we have:

\[ Z = \frac{1 - x^2 - 2ix}{1 + x^2}. \]

### Step 3: Separate real and imaginary parts

Now, we can separate the real and imaginary parts:

\[ Z = \frac{1 - x^2}{1 + x^2} - i \frac{2x}{1 + x^2}. \]

### Step 4: Compare with the right-hand side

We compare this with the right-hand side:

\[ \sin\left(\frac{\pi}{7}\right) - i \cos\left(\frac{\pi}{7}\right). \]

From this, we can equate the real and imaginary parts:

1. Real part:
\[ \frac{1 - x^2}{1 + x^2} = \sin\left(\frac{\pi}{7}\right). \]

2. Imaginary part:
\[ \frac{2x}{1 + x^2} = \cos\left(\frac{\pi}{7}\right). \]

### Step 5: Solve the equations

From the first equation, we can rearrange it:

\[ 1 - x^2 = \sin\left(\frac{\pi}{7}\right)(1 + x^2). \]

This simplifies to:

\[ 1 - x^2 = \sin\left(\frac{\pi}{7}\right) + \sin\left(\frac{\pi}{7}\right)x^2, \]

which leads to:

\[ (1 - \sin\left(\frac{\pi}{7}\right)) = x^2(1 + \sin\left(\frac{\pi}{7}\right)). \]

Thus, we have:

\[ x^2 = \frac{1 - \sin\left(\frac{\pi}{7}\right)}{1 + \sin\left(\frac{\pi}{7}\right)}. \]

### Step 6: Substitute \( x = \tan\left(\frac{k\pi}{28}\right) \)

Now substituting \( x = \tan\left(\frac{k\pi}{28}\right) \):

\[ \tan^2\left(\frac{k\pi}{28}\right) = \frac{1 - \sin\left(\frac{\pi}{7}\right)}{1 + \sin\left(\frac{\pi}{7}\right)}. \]

### Step 7: Use the double angle identity

Using the identity \( \tan^2(\theta) = \frac{1 - \cos(2\theta)}{\cos(2\theta)} \):

We can express \( \tan^2\left(\frac{k\pi}{28}\right) \) in terms of sine and cosine.

### Step 8: Find \( k \)

After solving the above equations, we can find the value of \( k \) that satisfies the original equation.

From the comparison of angles, we can deduce that:

\[ \frac{k\pi}{28} = \frac{5\pi}{28} \implies k = 5. \]

### Final Answer

Thus, the value of \( k \) can be:

\[ \boxed{5}. \]