The value of `2"cos"(pi)/(13)"cos"(9pi)/(13)+"cos"(3pi)/(13)+"cos"(5pi)/(13)` is

Options

A. `(1)/(2)`
B. 0
C. `-(1)/(2)`
D. `(1)/(8)`

Answer

To solve the problem, we need to find the value of the expression:

\[ 2 \cos\left(\frac{\pi}{13}\right) \cos\left(\frac{9\pi}{13}\right) + \cos\left(\frac{3\pi}{13}\right) + \cos\left(\frac{5\pi}{13}\right) \]

### Step 1: Substitute \( \theta = \frac{\pi}{13} \)

We can rewrite the expression in terms of \( \theta \):

\[ 2 \cos(\theta) \cos(9\theta) + \cos(3\theta) + \cos(5\theta) \]

### Step 2: Use the identity for \( 2 \cos A \cos B \)

Recall the identity:

\[ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \]

Applying this identity to \( 2 \cos(\theta) \cos(9\theta) \):

\[ 2 \cos(\theta) \cos(9\theta) = \cos(10\theta) + \cos(8\theta) \]

### Step 3: Rewrite the expression

Now substitute this back into the expression:

\[ \cos(10\theta) + \cos(8\theta) + \cos(3\theta) + \cos(5\theta) \]

### Step 4: Group the cosines

Notice that we can group the cosines:

\[ \cos(10\theta) + \cos(3\theta) + \cos(8\theta) + \cos(5\theta) \]

### Step 5: Use the property of cosine

Using the property that \( \cos(\pi - x) = -\cos(x) \), we can rewrite \( \cos(10\theta) \) and \( \cos(8\theta) \):

\[ \cos(10\theta) = \cos\left(\pi - 3\theta\right) = -\cos(3\theta) \]
\[ \cos(8\theta) = \cos\left(\pi - 5\theta\right) = -\cos(5\theta) \]

### Step 6: Substitute back into the expression

Now substitute these back into the expression:

\[ -\cos(3\theta) + \cos(3\theta) - \cos(5\theta) + \cos(5\theta) = 0 \]

### Conclusion

Thus, the value of the original expression is:

\[ \boxed{0} \]