Let `f_n(theta)=(cos theta/2+cos 2theta + cos (7theta)/2+...+cos (3n-2) (theta/2))/(sin theta/2+sin 2theta + sin (7theta)/2+....+ sin (3n-2)(theta/2))` then `f_3((3pi)/16)`

Options

A. `f_(3)((3pi)/(16))=sqrt(2)-1`
B. `f_(5)((pi)/(28))=sqrt(2)+1`
C. `f_(7)((pi)/(60))=(2+sqrt(3))`
D. none of these.

Explanation

To solve the given problem, we will follow these steps:

### Step 1: Rewrite the function
The function \( f_n(\theta) \) is given as:
\[ f_n(\theta) = \frac{\cos \frac{\theta}{2} + \cos 2\theta + \cos \frac{7\theta}{2} + \ldots + \cos \left( (3n-2) \frac{\theta}{2} \right)}{\sin \frac{\theta}{2} + \sin 2\theta + \sin \frac{7\theta}{2} + \ldots + \sin \left( (3n-2) \frac{\theta}{2} \right)} \]

### Step 2: Express as summations
We can express the numerator and denominator as summations:
\[ f_n(\theta) = \frac{\sum_{k=0}^{n-1} \cos \left( \frac{\theta}{2} + k \cdot \frac{3\theta}{2} \right)}{\sum_{k=0}^{n-1} \sin \left( \frac{\theta}{2} + k \cdot \frac{3\theta}{2} \right)} \]

### Step 3: Use summation formulas
Using the summation formulas for cosine and sine:
\[ \sum_{k=0}^{n-1} \cos (a + kd) = \frac{\sin \left( \frac{nd}{2} \right) \cdot \cos \left( a + \frac{(n-1)d}{2} \right)}{\sin \left( \frac{d}{2} \right)} \]
\[ \sum_{k=0}^{n-1} \sin (a + kd) = \frac{\sin \left( \frac{nd}{2} \right) \cdot \sin \left( a + \frac{(n-1)d}{2} \right)}{\sin \left( \frac{d}{2} \right)} \]

### Step 4: Substitute values
Here, we have:
- \( a = \frac{\theta}{2} \)
- \( d = \frac{3\theta}{2} \)
- \( n = 3 \)

Substituting these values into the formulas:
\[ \text{Numerator: } \sum_{k=0}^{2} \cos \left( \frac{\theta}{2} + k \cdot \frac{3\theta}{2} \right) = \frac{\sin \left( \frac{3 \cdot \frac{3\theta}{2}}{2} \right) \cdot \cos \left( \frac{\theta}{2} + \frac{2 \cdot \frac{3\theta}{2}}{2} \right)}{\sin \left( \frac{\frac{3\theta}{2}}{2} \right)} \]
\[ = \frac{\sin \left( \frac{9\theta}{4} \right) \cdot \cos \left( \frac{5\theta}{2} \right)}{\sin \left( \frac{3\theta}{4} \right)} \]

\[ \text{Denominator: } \sum_{k=0}^{2} \sin \left( \frac{\theta}{2} + k \cdot \frac{3\theta}{2} \right) = \frac{\sin \left( \frac{3 \cdot \frac{3\theta}{2}}{2} \right) \cdot \sin \left( \frac{\theta}{2} + \frac{2 \cdot \frac{3\theta}{2}}{2} \right)}{\sin \left( \frac{\frac{3\theta}{2}}{2} \right)} \]
\[ = \frac{\sin \left( \frac{9\theta}{4} \right) \cdot \sin \left( \frac{5\theta}{2} \right)}{\sin \left( \frac{3\theta}{4} \right)} \]

### Step 5: Simplify the function
Now, substituting back into \( f_n(\theta) \):
\[ f_3(\theta) = \frac{\frac{\sin \left( \frac{9\theta}{4} \right) \cdot \cos \left( \frac{5\theta}{2} \right)}{\sin \left( \frac{3\theta}{4} \right)}}{\frac{\sin \left( \frac{9\theta}{4} \right) \cdot \sin \left( \frac{5\theta}{2} \right)}{\sin \left( \frac{3\theta}{4} \right)}} \]
The \( \sin \left( \frac{9\theta}{4} \right) \) and \( \sin \left( \frac{3\theta}{4} \right) \) cancel out:
\[ f_3(\theta) = \frac{\cos \left( \frac{5\theta}{2} \right)}{\sin \left( \frac{5\theta}{2} \right)} = \cot \left( \frac{5\theta}{2} \right) \]

### Step 6: Evaluate for \( \theta = \frac{3\pi}{16} \)
Now, substituting \( \theta = \frac{3\pi}{16} \):
\[ f_3\left( \frac{3\pi}{16} \right) = \cot \left( \frac{5 \cdot \frac{3\pi}{16}}{2} \right) = \cot \left( \frac{15\pi}{32} \right) \]

### Final Answer
Thus, the value of \( f_3\left( \frac{3\pi}{16} \right) \) is:
\[ \cot \left( \frac{15\pi}{32} \right) \]