`cos "" (pi )/( 65) cos "" ( 2pi )/( 65)cos"" ( 4pi )/( 65) cos "" ( 8 pi )/(65)cos"" ( 16pi )/( 65) cos "" ( 32pi )/( 65)=`

Options

A. `1//8`
B. `1//16`
C. `1//32`
D. `1//64`

Answer

To solve the expression

\[ \cos\left(\frac{\pi}{65}\right) \cos\left(\frac{2\pi}{65}\right) \cos\left(\frac{4\pi}{65}\right) \cos\left(\frac{8\pi}{65}\right) \cos\left(\frac{16\pi}{65}\right) \cos\left(\frac{32\pi}{65}\right), \]

we can follow these steps:

### Step 1: Define A
Let \( A = \frac{\pi}{65} \). Then we can rewrite the expression as:

\[ \cos A \cos(2A) \cos(4A) \cos(8A) \cos(16A) \cos(32A). \]

### Step 2: Use the Cosine Product Formula
We can use the formula for the product of cosines:

\[ \cos A \cos(2A) \cos(4A) \cdots \cos(2^{n-1} A) = \frac{\sin(2^n A)}{2^n \sin A}. \]

In our case, we have \( n = 6 \) since we have 6 terms.

### Step 3: Substitute n into the Formula
Substituting \( n = 6 \) into the formula gives us:

\[ \cos A \cos(2A) \cos(4A) \cos(8A) \cos(16A) \cos(32A) = \frac{\sin(2^6 A)}{2^6 \sin A} = \frac{\sin(64A)}{64 \sin A}. \]

### Step 4: Substitute A back into the Equation
Now substituting \( A = \frac{\pi}{65} \):

\[ \sin(64A) = \sin\left(64 \cdot \frac{\pi}{65}\right) = \sin\left(\frac{64\pi}{65}\right). \]

### Step 5: Use the Sine Identity
Using the identity \( \sin(\pi - x) = \sin x \):

\[ \sin\left(\frac{64\pi}{65}\right) = \sin\left(\pi - \frac{\pi}{65}\right) = \sin\left(\frac{\pi}{65}\right). \]

### Step 6: Substitute back into the Expression
Now substituting this back into our expression:

\[ \frac{\sin\left(\frac{64\pi}{65}\right)}{64 \sin\left(\frac{\pi}{65}\right)} = \frac{\sin\left(\frac{\pi}{65}\right)}{64 \sin\left(\frac{\pi}{65}\right)}. \]

### Step 7: Simplify
The \( \sin\left(\frac{\pi}{65}\right) \) terms cancel out:

\[ \frac{1}{64}. \]

### Final Answer
Thus, the value of the original expression is:

\[ \frac{1}{64}. \]