The value of `sintheta + sin3theta +sin5theta + "….."+ sin(2n-1)theta` is

Options

A. `(cos^(2)ntheta)/(sintheta)`
B. `(sin^(2)ntheta)/(costheta)`
C. `(sin^(2)ntheta)/(sin theta)`
D. `(cos^(2)ntheta)/(costheta)`

Answer

To find the value of the expression \( \sin \theta + \sin 3\theta + \sin 5\theta + \ldots + \sin(2n-1)\theta \), we can use a formula for the sum of sine functions.

### Step-by-Step Solution:

1. **Identify the Series**: The series we need to evaluate is \( \sin \theta + \sin 3\theta + \sin 5\theta + \ldots + \sin(2n-1)\theta \). This can be expressed as:
\[ S = \sum_{k=0}^{n-1} \sin((2k+1)\theta) \]

2. **Use the Sine Sum Formula**: We can use the formula for the sum of sines:
\[ \sum_{k=0}^{n-1} \sin(a + kd) = \frac{\sin\left(\frac{nd}{2}\right)}{\sin\left(\frac{d}{2}\right)} \cdot \sin\left(a + \frac{(n-1)d}{2}\right) \]
In our case, \( a = \theta \) and \( d = 2\theta \).

3. **Substitute Values**: Plugging in the values into the formula:
\[ S = \frac{\sin\left(\frac{n \cdot 2\theta}{2}\right)}{\sin\left(\frac{2\theta}{2}\right)} \cdot \sin\left(\theta + \frac{(n-1) \cdot 2\theta}{2}\right) \]
Simplifying this gives:
\[ S = \frac{\sin(n\theta)}{\sin(\theta)} \cdot \sin\left(\theta + (n-1)\theta\right) \]

4. **Simplify Further**: The term \( \sin\left(\theta + (n-1)\theta\right) \) simplifies to \( \sin(n\theta) \):
\[ S = \frac{\sin(n\theta)}{\sin(\theta)} \cdot \sin(n\theta) \]

5. **Final Result**: Therefore, the final expression for the sum \( S \) is:
\[ S = \frac{\sin^2(n\theta)}{\sin(\theta)} \]

### Final Answer:
\[ \sin \theta + \sin 3\theta + \sin 5\theta + \ldots + \sin(2n-1)\theta = \frac{\sin^2(n\theta)}{\sin(\theta)} \]