`sin(pi/2^2009)cos(pi/2^2009)cos(pi/2^2008).....cos(pi/2^2)` is

Options

A. `(1)/(2^(2007))`
B. `(1)/(2^(2008))`
C. `(1)/(2^(2009))`
D. `(1)/(2^(2010))`

Answer

To solve the expression \( \sin\left(\frac{\pi}{2^{2009}}\right) \cos\left(\frac{\pi}{2^{2009}}\right) \cos\left(\frac{\pi}{2^{2008}}\right) \cdots \cos\left(\frac{\pi}{2^2}\right) \), we will use the identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \).

### Step-by-Step Solution:

1. **Define the Expression**:
Let
\[ y = \sin\left(\frac{\pi}{2^{2009}}\right) \cos\left(\frac{\pi}{2^{2009}}\right) \cos\left(\frac{\pi}{2^{2008}}\right) \cdots \cos\left(\frac{\pi}{2^2}\right). \]

2. **Use the Double Angle Identity**:
We know that
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta). \]
Therefore, we can rewrite \( \sin\left(\frac{\pi}{2^{2009}}\right) \cos\left(\frac{\pi}{2^{2009}}\right) \) as:
\[ \sin\left(\frac{\pi}{2^{2009}}\right) \cos\left(\frac{\pi}{2^{2009}}\right) = \frac{1}{2} \sin\left(\frac{\pi}{2^{2008}}\right). \]
Thus, we have:
\[ y = \frac{1}{2} \sin\left(\frac{\pi}{2^{2008}}\right) \cos\left(\frac{\pi}{2^{2008}}\right) \cos\left(\frac{\pi}{2^{2007}}\right) \cdots \cos\left(\frac{\pi}{2^2}\right). \]

3. **Repeat the Process**:
Now, we can apply the same identity to \( \sin\left(\frac{\pi}{2^{2008}}\right) \cos\left(\frac{\pi}{2^{2008}}\right) \):
\[ \sin\left(\frac{\pi}{2^{2008}}\right) \cos\left(\frac{\pi}{2^{2008}}\right) = \frac{1}{2} \sin\left(\frac{\pi}{2^{2007}}\right). \]
So now we have:
\[ y = \frac{1}{2} \cdot \frac{1}{2} \sin\left(\frac{\pi}{2^{2007}}\right) \cos\left(\frac{\pi}{2^{2007}}\right) \cos\left(\frac{\pi}{2^{2006}}\right) \cdots \cos\left(\frac{\pi}{2^2}\right). \]

4. **Continue Until the Last Term**:
Continuing this process, we will keep halving the expression and introducing sine terms until we reach:
\[ y = \frac{1}{2^{2008}} \sin\left(\frac{\pi}{2^2}\right). \]

5. **Evaluate the Final Sine Term**:
We know that:
\[ \sin\left(\frac{\pi}{2^2}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}. \]
Therefore, substituting this back into our expression gives:
\[ y = \frac{1}{2^{2008}} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2^{2009}}. \]

### Final Answer:
\[ y = \frac{\sqrt{2}}{2^{2009}}. \]