Let `f_n(a)=(sinalpha+sin3alpha+sin5alpha+...+sin(2n-1)alpha)/(cosalpha+cos3alpha+cos5alpha+...+cos(2n-1)alpha)` Then, the value of `f_4(pi/32)` is equal to

Options

A. `sqrt(2)+1`
B. `sqrt(2)-1`
C. `2+sqrt(3)`
D. `2-sqrt(3)`

Answer

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

\[ f_n(a) = \frac{\sin \alpha + \sin 3\alpha + \sin 5\alpha + \ldots + \sin(2n-1)\alpha}{\cos \alpha + \cos 3\alpha + \cos 5\alpha + \ldots + \cos(2n-1)\alpha} \]

For \( n = 4 \) and \( \alpha = \frac{\pi}{32} \), we have:

\[ f_4\left(\frac{\pi}{32}\right) = \frac{\sin\left(\frac{\pi}{32}\right) + \sin\left(\frac{3\pi}{32}\right) + \sin\left(\frac{5\pi}{32}\right) + \sin\left(\frac{7\pi}{32}\right)}{\cos\left(\frac{\pi}{32}\right) + \cos\left(\frac{3\pi}{32}\right) + \cos\left(\frac{5\pi}{32}\right) + \cos\left(\frac{7\pi}{32}\right)} \]

### Step 1: Combine Sine Terms
We can use the sine addition formula to combine the terms in the numerator:

\[ \sin a + \sin b = 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \]

We will pair the terms:

1. \(\sin\left(\frac{\pi}{32}\right) + \sin\left(\frac{7\pi}{32}\right)\)
2. \(\sin\left(\frac{3\pi}{32}\right) + \sin\left(\frac{5\pi}{32}\right)\)

Calculating the first pair:

\[ \sin\left(\frac{\pi}{32}\right) + \sin\left(\frac{7\pi}{32}\right) = 2 \sin\left(\frac{\frac{\pi}{32} + \frac{7\pi}{32}}{2}\right) \cos\left(\frac{\frac{7\pi}{32} - \frac{\pi}{32}}{2}\right) = 2 \sin\left(\frac{4\pi}{32}\right) \cos\left(\frac{3\pi}{32}\right) = 2 \sin\left(\frac{\pi}{8}\right) \cos\left(\frac{3\pi}{32}\right) \]

Calculating the second pair:

\[ \sin\left(\frac{3\pi}{32}\right) + \sin\left(\frac{5\pi}{32}\right) = 2 \sin\left(\frac{\frac{3\pi}{32} + \frac{5\pi}{32}}{2}\right) \cos\left(\frac{\frac{5\pi}{32} - \frac{3\pi}{32}}{2}\right) = 2 \sin\left(\frac{4\pi}{32}\right) \cos\left(\frac{\pi}{32}\right) = 2 \sin\left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{32}\right) \]

Combining both results:

\[ \text{Numerator} = 2 \sin\left(\frac{\pi}{8}\right) \left(\cos\left(\frac{3\pi}{32}\right) + \cos\left(\frac{\pi}{32}\right)\right) \]

### Step 2: Combine Cosine Terms
Now, we will combine the cosine terms in the denominator:

Using the same pairing:

1. \(\cos\left(\frac{\pi}{32}\right) + \cos\left(\frac{7\pi}{32}\right)\)
2. \(\cos\left(\frac{3\pi}{32}\right) + \cos\left(\frac{5\pi}{32}\right)\)

Calculating the first pair:

\[ \cos\left(\frac{\pi}{32}\right) + \cos\left(\frac{7\pi}{32}\right) = 2 \cos\left(\frac{\frac{\pi}{32} + \frac{7\pi}{32}}{2}\right) \cos\left(\frac{\frac{7\pi}{32} - \frac{\pi}{32}}{2}\right) = 2 \cos\left(\frac{4\pi}{32}\right) \cos\left(\frac{3\pi}{32}\right) = 2 \cos\left(\frac{\pi}{8}\right) \cos\left(\frac{3\pi}{32}\right) \]

Calculating the second pair:

\[ \cos\left(\frac{3\pi}{32}\right) + \cos\left(\frac{5\pi}{32}\right) = 2 \cos\left(\frac{\frac{3\pi}{32} + \frac{5\pi}{32}}{2}\right) \cos\left(\frac{\frac{5\pi}{32} - \frac{3\pi}{32}}{2}\right) = 2 \cos\left(\frac{4\pi}{32}\right) \cos\left(\frac{\pi}{32}\right) = 2 \cos\left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{32}\right) \]

Combining both results:

\[ \text{Denominator} = 2 \cos\left(\frac{\pi}{8}\right) \left(\cos\left(\frac{3\pi}{32}\right) + \cos\left(\frac{\pi}{32}\right)\right) \]

### Step 3: Simplifying the Function
Now, substituting back into \( f_4\left(\frac{\pi}{32}\right) \):

\[ f_4\left(\frac{\pi}{32}\right) = \frac{2 \sin\left(\frac{\pi}{8}\right) \left(\cos\left(\frac{3\pi}{32}\right) + \cos\left(\frac{\pi}{32}\right)\right)}{2 \cos\left(\frac{\pi}{8}\right) \left(\cos\left(\frac{3\pi}{32}\right) + \cos\left(\frac{\pi}{32}\right)\right)} \]

The terms \(\left(\cos\left(\frac{3\pi}{32}\right) + \cos\left(\frac{\pi}{32}\right)\right)\) cancel out:

\[ f_4\left(\frac{\pi}{32}\right) = \frac{\sin\left(\frac{\pi}{8}\right)}{\cos\left(\frac{\pi}{8}\right)} = \tan\left(\frac{\pi}{8}\right) \]

### Step 4: Finding the Value of \(\tan\left(\frac{\pi}{8}\right)\)
Using the half-angle identity:

\[ \tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1 \]

### Final Result
Thus, the value of \( f_4\left(\frac{\pi}{32}\right) \) is:

\[ \sqrt{2} - 1 \]