If `x+(1)/(x)=2costheta`, then `x^(n)+(1)/(x^(n))` is equal to

Options

A. `2cosntheta`
B. `2sinntheta`
C. `cosntheta`
D. `sinntheta`

Explanation

To solve the problem where \( x + \frac{1}{x} = 2 \cos \theta \) and we need to find \( x^n + \frac{1}{x^n} \), we can follow these steps:

### Step 1: Rewrite the given equation
We start with the equation:
\[ x + \frac{1}{x} = 2 \cos \theta \]

### Step 2: Multiply both sides by \( x \)
Multiplying both sides by \( x \) gives:
\[ x^2 + 1 = 2x \cos \theta \]

### Step 3: Rearrange into standard quadratic form
Rearranging the equation, we get:
\[ x^2 - 2x \cos \theta + 1 = 0 \]

### Step 4: Apply the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -2 \cos \theta, c = 1 \):
\[ x = \frac{2 \cos \theta \pm \sqrt{(2 \cos \theta)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ = \frac{2 \cos \theta \pm \sqrt{4 \cos^2 \theta - 4}}{2} \]
\[ = \frac{2 \cos \theta \pm 2 \sqrt{\cos^2 \theta - 1}}{2} \]
\[ = \cos \theta \pm \sqrt{\cos^2 \theta - 1} \]

### Step 5: Simplify the square root
Since \( \cos^2 \theta - 1 = -\sin^2 \theta \):
\[ = \cos \theta \pm i \sin \theta \]

### Step 6: Express \( x \) in exponential form
Thus, we can express \( x \) as:
\[ x = e^{i\theta} \quad \text{or} \quad x = e^{-i\theta} \]

### Step 7: Calculate \( x^n \)
Now, we can calculate \( x^n \):
\[ x^n = (e^{i\theta})^n = e^{in\theta} \quad \text{or} \quad x^n = (e^{-i\theta})^n = e^{-in\theta} \]

### Step 8: Calculate \( \frac{1}{x^n} \)
The reciprocal is:
\[ \frac{1}{x^n} = e^{-in\theta} \quad \text{or} \quad \frac{1}{x^n} = e^{in\theta} \]

### Step 9: Add \( x^n \) and \( \frac{1}{x^n} \)
Now, we add \( x^n \) and \( \frac{1}{x^n} \):
\[ x^n + \frac{1}{x^n} = e^{in\theta} + e^{-in\theta} \]
Using Euler's formula, this simplifies to:
\[ = 2 \cos(n\theta) \]

### Final Answer
Thus, we conclude that:
\[ x^n + \frac{1}{x^n} = 2 \cos(n\theta) \]

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