If `theta =(pi)/(2 ^(n) +1),` then
`cos theta cos 2 theta cos 2^(2) theta …cos 2 ^(n-1)theta ` is equal to

Options

A. `(1)/(2 ^(n))`
B. `cos theta `
C. `2`
D. `2^(n)`

Answer

To solve the problem, we need to find the value of the expression:

\[ \cos \theta \cos 2\theta \cos 2^2 \theta \ldots \cos 2^{n-1} \theta \]

where \(\theta = \frac{\pi}{2^n + 1}\).

### Step-by-step Solution:

1. **Substituting the value of \(\theta\)**:
We have:
\[ \theta = \frac{\pi}{2^n + 1} \]
Therefore, we can express \(2^n \theta\) as:
\[ 2^n \theta = 2^n \cdot \frac{\pi}{2^n + 1} = \frac{2^n \pi}{2^n + 1} \]

2. **Relating \(2^n \theta\) to \(\pi\)**:
We can rearrange this to find:
\[ 2^n \theta + \theta = \frac{2^n \pi}{2^n + 1} + \frac{\pi}{2^n + 1} = \frac{(2^n + 1)\pi}{2^n + 1} = \pi \]
Thus, we have:
\[ 2^n \theta = \pi - \theta \]

3. **Using the product formula**:
The product we need to evaluate can be expressed using a known trigonometric identity:
\[ \cos \theta \cos 2\theta \cos 4\theta \ldots \cos 2^{n-1}\theta = \frac{\sin 2^n \theta}{2^n \sin \theta} \]

4. **Substituting \(2^n \theta\)**:
From our earlier calculation, we have:
\[ 2^n \theta = \pi - \theta \]
Therefore:
\[ \sin 2^n \theta = \sin(\pi - \theta) = \sin \theta \]

5. **Putting it all together**:
Now substituting back into the product formula:
\[ \cos \theta \cos 2\theta \cos 4\theta \ldots \cos 2^{n-1}\theta = \frac{\sin 2^n \theta}{2^n \sin \theta} = \frac{\sin \theta}{2^n \sin \theta} \]
The \(\sin \theta\) terms cancel out:
\[ = \frac{1}{2^n} \]

### Final Result:
Thus, the value of the expression is:
\[ \frac{1}{2^n} \]