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Mathematics Question on Trigonometric Functions

The value of the sin⁡  1∘+sin⁡  2∘+......+sin⁡359∘\sin \; 1^{\circ} + \sin \; 2^{\circ} +......+ \sin 359^{\circ} is equal to

A

0

B

1

C

-1

D

180180

Answer

0

Explanation

Solution

sin⁡1∘+sin⁡2∘+sin⁡3∘+…+sin⁡359∘\sin 1^{\circ}+\sin 2^{\circ}+\sin 3^{\circ}+\ldots+\sin 359^{\circ}
=(sin⁡1∘+sin⁡359∘)+(sin⁡2∘+sin⁡358∘)+…=\left(\sin 1^{\circ}+\sin 359^{\circ}\right)+\left(\sin 2^{\circ}+\sin 358^{\circ}\right)+\ldots
+sin⁡180∘+\sin 180^{\circ}
=0+0+0+…+0=0=0+0+0+\ldots+0=0
[∵sin⁡θ+sin⁡(360∘−θ)=0]\left[\because \sin \theta+\sin \left(360^{\circ}-\theta\right)=0\right]