If `x^2-2xcos theta+1=0`, then the value of `x^(2n)-2x^n cosntheta+1, n in N` is equal to

Options

A. `cos 2n theta`
B. `sin 2n theta`
C. 0
D. `cos ntheta`

Answer

To solve the problem, we need to analyze the given quadratic equation and derive the required expression step by step.

### Step-by-Step Solution:

1. **Given Equation**:
We start with the quadratic equation:
\[ x^2 - 2x \cos \theta + 1 = 0 \]

2. **Finding Roots**:
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -2\cos\theta \), and \( c = 1 \).
\[ x = \frac{2\cos\theta \pm \sqrt{(-2\cos\theta)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ = \frac{2\cos\theta \pm \sqrt{4\cos^2\theta - 4}}{2} \]
\[ = \frac{2\cos\theta \pm 2\sqrt{\cos^2\theta - 1}}{2} \]
\[ = \cos\theta \pm \sqrt{\cos^2\theta - 1} \]

3. **Simplifying the Square Root**:
Since \( \cos^2\theta - 1 = -\sin^2\theta \), we can rewrite the roots as:
\[ x = \cos\theta \pm i\sin\theta \]
This can be expressed using Euler's formula:
\[ x = e^{i\theta} \quad \text{or} \quad x = e^{-i\theta} \]

4. **Finding \( x^n \)**:
Now we need to compute \( x^n \):
\[ x^n = (e^{i\theta})^n = e^{in\theta} \quad \text{or} \quad x^n = (e^{-i\theta})^n = e^{-in\theta} \]

5. **Substituting into the Expression**:
We need to compute:
\[ x^{2n} - 2x^n \cos n\theta + 1 \]
Substituting \( x^n = e^{in\theta} \):
\[ x^{2n} = (e^{in\theta})^2 = e^{2in\theta} \]
Thus, the expression becomes:
\[ e^{2in\theta} - 2e^{in\theta} \cos n\theta + 1 \]

6. **Using Euler's Formula**:
We can express \( e^{2in\theta} \) as:
\[ e^{2in\theta} = \cos(2n\theta) + i\sin(2n\theta) \]
And for \( -2e^{in\theta} \cos n\theta \):
\[ -2e^{in\theta} \cos n\theta = -2(\cos(n\theta) + i\sin(n\theta)) \cos n\theta \]
This simplifies to:
\[ -2\cos^2(n\theta) - 2i\sin(n\theta)\cos(n\theta) \]

7. **Combining Terms**:
Now combining all terms:
\[ (\cos(2n\theta) - 2\cos^2(n\theta) + 1) + i(\sin(2n\theta) - 2\sin(n\theta)\cos(n\theta)) \]

8. **Using Trigonometric Identities**:
We can use the identity \( \cos(2n\theta) = 2\cos^2(n\theta) - 1 \):
\[ 2\cos^2(n\theta) - 2\cos^2(n\theta) + 1 = 0 \]
The imaginary part also simplifies to zero:
\[ \sin(2n\theta) - 2\sin(n\theta)\cos(n\theta) = 0 \]

9. **Final Result**:
Thus, the entire expression simplifies to:
\[ 0 \]

### Conclusion:
The value of \( x^{2n} - 2x^n \cos n\theta + 1 \) is equal to \( 0 \).