If `x+1/x=2 cos theta`, then the value of `x^n+1/x^n`

Answer

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

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

### Step 2: Use the double angle identity
We can simplify \( 4 \cos^2 \theta - 2 \) using the double angle identity:
\[ 4 \cos^2 \theta - 2 = 2(2 \cos^2 \theta - 1) = 2 \cos 2\theta \]
Thus, we have:
\[ x^2 + \frac{1}{x^2} = 2 \cos 2\theta \]

### Step 3: Cube the original equation
Now, we will cube the original equation:
\[ \left( x + \frac{1}{x} \right)^3 = (2 \cos \theta)^3 \]
Using the cube expansion formula:
\[ x^3 + 3\left( x + \frac{1}{x} \right) + \frac{1}{x^3} = 8 \cos^3 \theta \]
Substituting \( x + \frac{1}{x} = 2 \cos \theta \):
\[ x^3 + \frac{1}{x^3} + 6 \cos \theta = 8 \cos^3 \theta \]
Rearranging gives:
\[ x^3 + \frac{1}{x^3} = 8 \cos^3 \theta - 6 \cos \theta \]

### Step 4: Use the triple angle identity
We can express \( 8 \cos^3 \theta - 6 \cos \theta \) using the triple angle formula:
\[ 8 \cos^3 \theta - 6 \cos \theta = 2 \cos 3\theta \]
Thus, we have:
\[ x^3 + \frac{1}{x^3} = 2 \cos 3\theta \]

### Step 5: Generalize to \( n \)
From the pattern observed, we can generalize that:
\[ x^n + \frac{1}{x^n} = 2 \cos n\theta \]
This leads us to the final result:
\[ \boxed{2 \cos n\theta} \]