The equation whose roots are nth power of the roots of the equations is ` x^(2) - 2 x cos theta + 1 = 0` is given by

Options

A. `( x + cos n theta)^(2) + sin^(2) n theta = 0 `
B. ` ( x - cos n theta)^(2) + sin theta ^(2) n theta = 0 `
C. ` x^(2) + 2 x cos n theta + 1 = 0 `
D. ` x^(2) - 2 x cos n theta + 1 = 0 `

Answer

To solve the problem, we need to find the equation whose roots are the nth powers of the roots of the given equation \( x^2 - 2x \cos \theta + 1 = 0 \).

### Step-by-Step Solution:

1. **Identify the given quadratic equation**:
The given equation is:
\[ x^2 - 2x \cos \theta + 1 = 0 \]

2. **Find the roots of the quadratic equation**:
We will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2 \cos \theta \), and \( c = 1 \).

Plugging in the values:
\[ x = \frac{2 \cos \theta \pm \sqrt{(-2 \cos \theta)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ = \frac{2 \cos \theta \pm \sqrt{4 \cos^2 \theta - 4}}{2} \]
\[ = \frac{2 \cos \theta \pm 2 \sqrt{\cos^2 \theta - 1}}{2} \]
\[ = \cos \theta \pm \sqrt{\cos^2 \theta - 1} \]
Since \( \cos^2 \theta - 1 = -\sin^2 \theta \), we have:
\[ = \cos \theta \pm i \sin \theta \]
Thus, the roots are:
\[ z_1 = \cos \theta + i \sin \theta \quad \text{and} \quad z_2 = \cos \theta - i \sin \theta \]

3. **Find the nth powers of the roots**:
The nth powers of the roots are:
\[ z_1^n = (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) \]
\[ z_2^n = (\cos \theta - i \sin \theta)^n = \cos(n\theta) - i \sin(n\theta) \]

4. **Form the new quadratic equation**:
The new quadratic equation whose roots are \( z_1^n \) and \( z_2^n \) can be formed using the sum and product of the roots:
- Sum of the roots:
\[ z_1^n + z_2^n = (\cos(n\theta) + i \sin(n\theta)) + (\cos(n\theta) - i \sin(n\theta)) = 2 \cos(n\theta) \]
- Product of the roots:
\[ z_1^n \cdot z_2^n = (\cos(n\theta) + i \sin(n\theta))(\cos(n\theta) - i \sin(n\theta)) = \cos^2(n\theta) + \sin^2(n\theta) = 1 \]

Therefore, the quadratic equation is:
\[ x^2 - (z_1^n + z_2^n)x + z_1^n z_2^n = 0 \]
Plugging in the values:
\[ x^2 - 2 \cos(n\theta) x + 1 = 0 \]

5. **Final result**:
The required equation is:
\[ x^2 - 2 \cos(n\theta) x + 1 = 0 \]