Question

The value of $sin{10}^0+sin{20}^0+sin{30}^0+....+sin{360}^0$ is equal to-

• A. $0$
• B. $1$
• C. $\sqrt{3}$
• D. $2$

Answer 1

The correct option is A $0$

Value of $\sin {10}^o+\sin {20}^o+...\dots +\sin {360}^o$

using identity of series sum:

$\sin \alpha +\sin (\alpha +\beta )+\sin (\alpha +2\beta )+...\dots +\sin (\alpha +(n-1\left)\beta \right)$

$=\frac{\sin (2+\left(\frac{n-1}{2}\right)\beta )\sin \left(\frac{n}{2}\beta \right)}{\sin \left(\frac{\beta }{2}\right)}$

The above series have $36$ terms $\alpha ={10}^o$ and $\beta ={10}^o$

$\Rightarrow$Sum $=\frac{\sin (10+\left(\frac{35}{2}\right){10}^o)\sin \left(\left(\frac{36}{2}\right){10}^o\right)}{\sin \left(\frac{{10}^o}{2}\right)}$

$=\frac{\sin \left({185}^o\right)\sin \left({180}^o\right)}{\sin 5^o}=0$.