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

Options

A. `2sinn theta`
B. `2cosntheta`
C. `sin(2ntheta)`
D. `cos(2ntheta)`

Explanation

To solve the problem, we start with the given equation:

**Given:**
\[ x + \frac{1}{x} = 2 \cos \theta \]

We need to find:
\[ x^n + \frac{1}{x^n} \]

### Step 1: Square the given equation
We square both sides of the equation:
\[ \left( x + \frac{1}{x} \right)^2 = (2 \cos \theta)^2 \]
This gives us:
\[ x^2 + 2 + \frac{1}{x^2} = 4 \cos^2 \theta \]
Thus,
\[ x^2 + \frac{1}{x^2} = 4 \cos^2 \theta - 2 \]

### Step 2: Simplify using the double angle formula
We can factor out 2 from the right-hand side:
\[ x^2 + \frac{1}{x^2} = 2(2 \cos^2 \theta - 1) \]
Using the double angle formula \( \cos 2\theta = 2 \cos^2 \theta - 1 \), we rewrite it as:
\[ x^2 + \frac{1}{x^2} = 2 \cos 2\theta \]

### Step 3: Cube the original equation
Next, we cube the original equation:
\[ \left( x + \frac{1}{x} \right)^3 = (2 \cos \theta)^3 \]
This expands to:
\[ x^3 + \frac{1}{x^3} + 3\left( x + \frac{1}{x} \right) = 8 \cos^3 \theta \]
Substituting \( x + \frac{1}{x} = 2 \cos \theta \):
\[ x^3 + \frac{1}{x^3} + 3(2 \cos \theta) = 8 \cos^3 \theta \]
Thus,
\[ x^3 + \frac{1}{x^3} = 8 \cos^3 \theta - 6 \cos \theta \]

### Step 4: Recognize the pattern
From the calculations, we observe a pattern:
- For \( n = 1 \): \( x + \frac{1}{x} = 2 \cos \theta \)
- For \( n = 2 \): \( x^2 + \frac{1}{x^2} = 2 \cos 2\theta \)
- For \( n = 3 \): \( x^3 + \frac{1}{x^3} = 2 \cos 3\theta \)

### Step 5: Generalize for \( n \)
From the pattern we see:
\[ x^n + \frac{1}{x^n} = 2 \cos n\theta \]

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

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