Question

$\sin10^{\circ} +\sin 20^{\circ }+\sin 30^{\circ }+...+\sin360^{\circ } =$

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

Answer 1

Answer: (A)

$\sin10^{\circ} +\sin20^{\circ }+\sin30^{\circ}+...+\sin360^{\circ }$
$ =\sin10^{\circ }+\sin20^{\circ }+........+\sin180^{\circ }+ \sin190^{\circ }+...+\sin 360^{\circ } $
$= \left(\sin10^{\circ} +\sin20^{\circ }+...+\sin180^{\circ }\right)+\left(\sin190^{\circ }+ \sin200^{\circ }+.....+\sin360^{\circ }\right) $
$= \left(\sin10^{\circ }+\sin20^{\circ } +\sin180^{\circ } \right)+\left[\sin\left(180^{\circ }+ 10^{\circ }\right)+\sin\left(180^{\circ }+ 20^{\circ}\right)+....+\sin\left(180^{\circ} +180^{\circ}\right)\right]$
$= \left(\sin 10^{\circ } +\sin 20^{\circ } +...+ \sin180^{\circ }\right)+\left[-\sin 10^{\circ } -\sin 20^{\circ } - ..... -\sin180^{\circ }\right] = 0$