`sin 10^(@) + sin 20^(@) + sin30^(@) +…+ sin 360^(@) ` is equal to

Options

A. 0
B. 1
C. `-1`
D. `1/2`

Answer

To solve the problem \( \sin 10^\circ + \sin 20^\circ + \sin 30^\circ + \ldots + \sin 360^\circ \), we can use the properties of the sine function, particularly the periodicity and symmetry of the sine function.

### Step-by-step Solution:

1. **Identify the series**: The series consists of sine values from \( \sin 10^\circ \) to \( \sin 360^\circ \) in increments of \( 10^\circ \). Thus, we can express the series as:
\[ S = \sin 10^\circ + \sin 20^\circ + \sin 30^\circ + \ldots + \sin 360^\circ \]

2. **Group the terms**: Notice that sine has a periodicity of \( 360^\circ \) and the property \( \sin(360^\circ - \theta) = -\sin(\theta) \). We can pair the terms:
- \( \sin 10^\circ \) and \( \sin 350^\circ \)
- \( \sin 20^\circ \) and \( \sin 340^\circ \)
- \( \sin 30^\circ \) and \( \sin 330^\circ \)
- ...
- \( \sin 170^\circ \) and \( \sin 190^\circ \)
- \( \sin 180^\circ \)

3. **Calculate the pairs**: Each pair can be simplified:
\[ \sin 10^\circ + \sin 350^\circ = \sin 10^\circ - \sin 10^\circ = 0 \]
\[ \sin 20^\circ + \sin 340^\circ = \sin 20^\circ - \sin 20^\circ = 0 \]
Continuing this way, we find that:
\[ \sin 30^\circ + \sin 330^\circ = \sin 30^\circ - \sin 30^\circ = 0 \]
\[ \sin 40^\circ + \sin 320^\circ = \sin 40^\circ - \sin 40^\circ = 0 \]
This pattern continues until:
\[ \sin 170^\circ + \sin 190^\circ = \sin 170^\circ - \sin 170^\circ = 0 \]

4. **Consider the middle term**: The only term that does not have a pair is \( \sin 180^\circ \), which is equal to 0.

5. **Combine all results**: Since all pairs sum to zero and the middle term is also zero, we conclude:
\[ S = 0 \]

### Final Answer:
Thus, the value of \( \sin 10^\circ + \sin 20^\circ + \sin 30^\circ + \ldots + \sin 360^\circ \) is:
\[ \boxed{0} \]