The equation whose roots are `nth` power of the roots of the equation, `x^2-2xcosphi + 1=0` is given by

Options

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

Answer

To find the equation whose roots are the nth powers of the roots of the equation \(x^2 - 2x \cos \phi + 1 = 0\), we will follow these steps:

### Step 1: Find the Roots of the Given Equation

The given equation is:
\[ x^2 - 2x \cos \phi + 1 = 0 \]
We will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -2 \cos \phi\), and \(c = 1\).

Substituting these values into the formula:
\[ x = \frac{2 \cos \phi \pm \sqrt{(-2 \cos \phi)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ = \frac{2 \cos \phi \pm \sqrt{4 \cos^2 \phi - 4}}{2} \]
\[ = \frac{2 \cos \phi \pm 2 \sqrt{\cos^2 \phi - 1}}{2} \]
\[ = \cos \phi \pm \sqrt{\cos^2 \phi - 1} \]
Since \(\cos^2 \phi - 1 = -\sin^2 \phi\), we can write:
\[ = \cos \phi \pm i \sin \phi \]
Thus, the roots are:
\[ e^{i\phi} \quad \text{and} \quad e^{-i\phi} \]

### Step 2: Find the nth Powers of the Roots

The nth powers of the roots are:
\[ (e^{i\phi})^n = e^{i n \phi} \quad \text{and} \quad (e^{-i\phi})^n = e^{-i n \phi} \]

### Step 3: Calculate the Sum and Product of the New Roots

**Sum of the new roots:**
\[ e^{i n \phi} + e^{-i n \phi} = 2 \cos(n \phi) \]

**Product of the new roots:**
\[ e^{i n \phi} \cdot e^{-i n \phi} = e^{0} = 1 \]

### Step 4: Formulate the New Quadratic Equation

Using the sum and product of the roots, we can write the new quadratic equation in the form:
\[ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \]
Substituting the values we found:
\[ x^2 - 2 \cos(n \phi) x + 1 = 0 \]

### Final Equation

Thus, the equation whose roots are the nth powers of the roots of the original equation is:
\[ x^2 - 2x \cos(n \phi) + 1 = 0 \]