If `theta=(2pi)/(2009), then costhetacos 2thetacos3theta... cos1004theta` is

Options

A. 0
B. `(1)/(2^(2008))`
C. `(1)/(2^(1004))`
D. `-(1)/(2^(1004))`

Explanation

To solve the problem, we need to evaluate the product \( P = \cos \theta \cos 2\theta \cos 3\theta \ldots \cos 1004\theta \) where \( \theta = \frac{2\pi}{2009} \).

### Step-by-Step Solution:

1. **Define the Product:**
Let \( P = \cos \theta \cos 2\theta \cos 3\theta \ldots \cos 1004\theta \).

2. **Introduce a Sine Product:**
Define another product \( Q = \sin \theta \sin 2\theta \sin 3\theta \ldots \sin 1004\theta \).

3. **Multiply \( P \) and \( Q \):**
Consider the product \( PQ = (\cos \theta \sin \theta)(\cos 2\theta \sin 2\theta)(\cos 3\theta \sin 3\theta) \ldots (\cos 1004\theta \sin 1004\theta) \).

4. **Use the Identity:**
Recall the identity \( 2 \sin x \cos x = \sin(2x) \). Thus, we can rewrite our product:
\[ PQ = \frac{1}{2^{1004}} \left( \sin(2\theta) \sin(4\theta) \sin(6\theta) \ldots \sin(2008\theta) \right) \]

5. **Evaluate the Sine Terms:**
Since \( \theta = \frac{2\pi}{2009} \), we have:
\[ 2008\theta = 2008 \cdot \frac{2\pi}{2009} = \frac{4016\pi}{2009} \]
The sine function has a period of \( 2\pi \), so we can reduce \( 2008\theta \) modulo \( 2\pi \):
\[ 2008\theta = 2\pi - \frac{2\pi}{2009} = 2\pi(1 - \frac{1}{2009}) \]
Thus, \( \sin(2008\theta) = \sin(2\pi - \frac{2\pi}{2009}) = \sin(\frac{2\pi}{2009} ) = \sin(\theta) \).

6. **Observe the Terms:**
The terms \( \sin(2\theta), \sin(4\theta), \ldots, \sin(2008\theta) \) will yield \( 0 \) at multiples of \( \pi \). Since \( \sin(k\theta) \) will be \( 0 \) for \( k = 2009 \), we find that \( PQ = 0 \).

7. **Conclusion:**
Since \( PQ = 0 \), it follows that either \( P = 0 \) or \( Q = 0 \). However, since \( \sin k\theta \) will be zero for some \( k \), we conclude that \( P \) must also be \( 0 \).

### Final Answer:
Thus, the value of \( P = \cos \theta \cos 2\theta \cos 3\theta \ldots \cos 1004\theta \) is **0**.