The value of `sin pi/n + sin (3pi)/n+ sin (5pi)/n+...` to n terms is equal to

Options

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

Answer

To find the value of the series \( S = \sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{3\pi}{n}\right) + \sin\left(\frac{5\pi}{n}\right) + \ldots + \sin\left(\frac{(2n-1)\pi}{n}\right) \), we will follow these steps:

### Step 1: Identify the series
The series consists of sine functions with arguments that are odd multiples of \( \frac{\pi}{n} \). The last term in the series is \( \sin\left(\frac{(2n-1)\pi}{n}\right) \).

### Step 2: Write the series in summation form
We can express the series as:
\[ S = \sum_{k=0}^{n-1} \sin\left(\frac{(2k+1)\pi}{n}\right) \]

### Step 3: Use the sine addition formula
We can pair the terms in the series:
\[ S = \left(\sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{(2n-1)\pi}{n}\right)\right) + \left(\sin\left(\frac{3\pi}{n}\right) + \sin\left(\frac{(2n-3)\pi}{n}\right)\right) + \ldots \]
Each pair can be simplified using the sine addition formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

### Step 4: Apply the formula to the first and last terms
For the first and last terms:
\[ A = \frac{\pi}{n}, \quad B = \frac{(2n-1)\pi}{n} \]
Calculating \( A + B \) and \( A - B \):
\[ A + B = \frac{\pi}{n} + \frac{(2n-1)\pi}{n} = \frac{2n\pi}{n} = 2\pi \]
\[ A - B = \frac{\pi}{n} - \frac{(2n-1)\pi}{n} = \frac{- (2n-2)\pi}{n} = -\frac{2(n-1)\pi}{n} \]
Thus:
\[ \sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{(2n-1)\pi}{n}\right) = 2 \sin\left(\pi\right) \cos\left(-\frac{(n-1)\pi}{n}\right) \]
Since \( \sin(\pi) = 0 \), this term becomes \( 0 \).

### Step 5: Repeat for all pairs
Following the same logic for all pairs, we find that each pair sums to zero. The last term \( \sin(\pi) \) also equals zero.

### Conclusion
Thus, the entire sum \( S \) evaluates to:
\[ S = 0 \]

### Final Answer
The value of the series \( \sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{3\pi}{n}\right) + \sin\left(\frac{5\pi}{n}\right) + \ldots + \sin\left(\frac{(2n-1)\pi}{n}\right) \) is \( 0 \).

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