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Mathematics Question on Trigonometric Functions

sin⁡10∘+sin⁡20∘+sin⁡30∘+...+sin⁡360∘=\sin10^{\circ} +\sin 20^{\circ }+\sin 30^{\circ }+...+\sin360^{\circ } =

A

00

B

11

C

3\sqrt{3}

D

22

Answer

00

Explanation

Solution

sin⁡10∘+sin⁡20∘+sin⁡30∘+...+sin⁡360∘\sin10^{\circ} +\sin20^{\circ }+\sin30^{\circ}+...+\sin360^{\circ }
=sin⁡10∘+sin⁡20∘+........+sin⁡180∘+sin⁡190∘+...+sin⁡360∘=\sin10^{\circ }+\sin20^{\circ }+........+\sin180^{\circ }+ \sin190^{\circ }+...+\sin 360^{\circ }
=(sin⁡10∘+sin⁡20∘+...+sin⁡180∘)+(sin⁡190∘+sin⁡200∘+.....+sin⁡360∘)= \left(\sin10^{\circ} +\sin20^{\circ }+...+\sin180^{\circ }\right)+\left(\sin190^{\circ }+ \sin200^{\circ }+.....+\sin360^{\circ }\right)
=(sin⁡10∘+sin⁡20∘+sin⁡180∘)+[sin⁡(180∘+10∘)+sin⁡(180∘+20∘)+....+sin⁡(180∘+180∘)]= \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]
=(sin⁡10∘+sin⁡20∘+...+sin⁡180∘)+[−sin⁡10∘−sin⁡20∘−.....−sin⁡180∘]=0= \left(\sin 10^{\circ } +\sin 20^{\circ } +...+ \sin180^{\circ }\right)+\left[-\sin 10^{\circ } -\sin 20^{\circ } - ..... -\sin180^{\circ }\right] = 0