The value of `cos(pi/4)*cos(pi/8)*cos(pi/16)....cos(pi/(2^n))` equals

Options

A. `1/2^(n)" cosec "(pi/(2n))`
B. `1/(2^(n-1) "cosec" (pi/(2^(n-1)))`
C. `1/2^(n)" cosec "(pi/(2^(n-1)))`
D. `1/(2^(n-1)) " cosec " (pi/2n)`

Answer

To solve the problem of finding the value of the expression \( \cos\left(\frac{\pi}{4}\right) \cdot \cos\left(\frac{\pi}{8}\right) \cdot \cos\left(\frac{\pi}{16}\right) \cdots \cos\left(\frac{\pi}{2^n}\right) \), we can follow these steps:

### Step 1: Write the expression
We start with the expression:
\[ E = \cos\left(\frac{\pi}{4}\right) \cdot \cos\left(\frac{\pi}{8}\right) \cdot \cos\left(\frac{\pi}{16}\right) \cdots \cos\left(\frac{\pi}{2^n}\right) \]

### Step 2: Multiply and divide by \( 2 \sin\left(\frac{\pi}{2^n}\right) \)
To simplify the expression, we multiply and divide by \( 2 \sin\left(\frac{\pi}{2^n}\right) \):
\[ E = \frac{E \cdot 2 \sin\left(\frac{\pi}{2^n}\right)}{2 \sin\left(\frac{\pi}{2^n}\right)} \]

### Step 3: Rearranging the terms
Rearranging gives us:
\[ E = \frac{2 \sin\left(\frac{\pi}{2^n}\right) \cdot \cos\left(\frac{\pi}{4}\right) \cdots \cos\left(\frac{\pi}{2^n}\right)}{2 \sin\left(\frac{\pi}{2^n}\right)} \]

### Step 4: Use the identity \( \sin(2a) = 2 \sin(a) \cos(a) \)
Using the identity \( \sin(2a) = 2 \sin(a) \cos(a) \), we can express the product of sine and cosine:
\[ 2 \sin\left(\frac{\pi}{2^n}\right) \cdot \cos\left(\frac{\pi}{2^n}\right) = \sin\left(\frac{\pi}{2^{n-1}}\right) \]
Continuing this process, we can keep applying the identity.

### Step 5: Continue applying the identity
Continuing this process, we will eventually reduce the expression down to:
\[ E = \frac{\sin\left(\frac{\pi}{2}\right)}{2^{n-1}} \]

### Step 6: Evaluate \( \sin\left(\frac{\pi}{2}\right) \)
Since \( \sin\left(\frac{\pi}{2}\right) = 1 \), we have:
\[ E = \frac{1}{2^{n-1}} \]

### Final Result
Thus, the value of the expression \( \cos\left(\frac{\pi}{4}\right) \cdot \cos\left(\frac{\pi}{8}\right) \cdots \cos\left(\frac{\pi}{2^n}\right) \) is:
\[ \frac{1}{2^{n-1}} \]