If the value of `"cos"pi/15"cos"(2pi)/15"cos"(3pi)/15"cos"(4pi)/15"cos"(5pi)/15"cos"(6pi)/15"cos"(7pi)/15` is `(1)/(2^(k))`, then which of the following is true ?

Options

A. k is prime
B. factors of k are 1,7
C. k is odd
D. k is even

Answer

To solve the problem, we need to evaluate the product:

\[ \cos\left(\frac{\pi}{15}\right) \cos\left(\frac{2\pi}{15}\right) \cos\left(\frac{3\pi}{15}\right) \cos\left(\frac{4\pi}{15}\right) \cos\left(\frac{5\pi}{15}\right) \cos\left(\frac{6\pi}{15}\right) \cos\left(\frac{7\pi}{15}\right) \]

and express it in the form \(\frac{1}{2^k}\).

### Step 1: Use the product-to-sum identities

We can utilize the identity \(2 \sin a \cos a = \sin(2a)\) to combine the cosine terms. We start with the first two terms:

\[ \cos\left(\frac{\pi}{15}\right) = \frac{\sin\left(\frac{2\pi}{15}\right)}{2\sin\left(\frac{\pi}{15}\right)} \]

Thus, we can rewrite the product as:

\[ \frac{\sin\left(\frac{2\pi}{15}\right)}{2\sin\left(\frac{\pi}{15}\right)} \cdot \cos\left(\frac{2\pi}{15}\right) \cdots \cos\left(\frac{7\pi}{15}\right) \]

### Step 2: Continue combining terms

Next, we apply the same identity to the next pair:

\[ \cos\left(\frac{2\pi}{15}\right) = \frac{\sin\left(\frac{4\pi}{15}\right)}{2\sin\left(\frac{2\pi}{15}\right)} \]

Now we can combine these terms:

\[ \frac{\sin\left(\frac{2\pi}{15}\right) \sin\left(\frac{4\pi}{15}\right)}{4\sin\left(\frac{\pi}{15}\right)\sin\left(\frac{2\pi}{15}\right)} \cdots \cos\left(\frac{7\pi}{15}\right) \]

### Step 3: Repeat the process

Continuing this process, we can keep combining the sine and cosine terms until we reach:

\[ \frac{\sin\left(\frac{12\pi}{15}\right)}{2^7 \sin\left(\frac{\pi}{15}\right)} \]

### Step 4: Simplify the sine term

Using the property \(\sin(\pi - x) = \sin(x)\):

\[ \sin\left(\frac{12\pi}{15}\right) = \sin\left(\frac{3\pi}{15}\right) = \sin\left(\frac{\pi}{5}\right) \]

### Step 5: Final expression

Thus, the product simplifies to:

\[ \frac{\sin\left(\frac{\pi}{5}\right)}{2^7 \sin\left(\frac{\pi}{15}\right)} \]

### Step 6: Set the equation

We know that the product equals \(\frac{1}{2^k}\), so we can equate:

\[ \frac{1}{2^7} = \frac{1}{2^k} \]

From this, we find:

\[ k = 7 \]

### Step 7: Conclusion

Now we can analyze the properties of \(k\):

1. **Is \(k\) prime?** Yes, 7 is a prime number.
2. **Is \(k\) odd?** Yes, 7 is odd.
3. **Is \(k\) even?** No, 7 is not even.

Thus, the correct statements about \(k\) are that it is prime and odd.