The value of `cosA cos2A cos2^2A......cos(2^(n-1)A)`, where `A in R` may be . (A) 1 (B) `-1` (C) 2 (D) `sin2^n A/(2^n sinA)`

Options

A. 1
B. 2
C. `-1`
D. `(sin2^(n)A)/(2n^(n)sinA)`

Answer

To find the value of \( \cos A \cos 2A \cos 2^2 A \ldots \cos(2^{n-1} A) \), we can follow these steps:

### Step-by-Step Solution:

1. **Expression Setup:**
We start with the product:
\[ P = \cos A \cos 2A \cos 2^2 A \ldots \cos(2^{n-1} A) \]

2. **Multiply and Divide by \( 2 \sin A \):**
To facilitate simplification, we multiply and divide by \( 2 \sin A \):
\[ P = \frac{\cos A \cos 2A \cos 2^2 A \ldots \cos(2^{n-1} A) \cdot 2 \sin A}{2 \sin A} \]

3. **Using the Identity \( 2 \sin x \cos x = \sin(2x) \):**
The numerator can be grouped:
\[ 2 \sin A \cos A = \sin(2A) \]
Thus, we can rewrite the product:
\[ P = \frac{\sin(2A) \cos 2A \cos 2^2 A \ldots \cos(2^{n-1} A)}{2 \sin A} \]

4. **Repeat the Process:**
We can repeat this process for \( \sin(2A) \):
\[ \sin(2A) \cos 2A = \frac{1}{2} \sin(4A) \]
So we can rewrite:
\[ P = \frac{\frac{1}{2} \sin(4A) \cos 2^2 A \ldots \cos(2^{n-1} A)}{2 \sin A} \]

5. **Continue Until the Last Term:**
Continuing this process, we will eventually get:
\[ P = \frac{\sin(2^n A)}{2^n \sin A} \]

6. **Final Result:**
Thus, the final expression for \( P \) is:
\[ P = \frac{\sin(2^n A)}{2^n \sin A} \]

### Conclusion:
The value of \( \cos A \cos 2A \cos 2^2 A \ldots \cos(2^{n-1} A) \) is:
\[ \frac{\sin(2^n A)}{2^n \sin A} \]

### Answer:
The correct option is \( \frac{\sin(2^n A)}{2^n \sin A} \), which corresponds to option (D).