sin10°+sin20°+sin30°+....+sin360°=?

Ans given is 0

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Solution

We know that sin(360-x)= -(sinx) and sin360=sin180=0

Now
sin10°+sin20°+sin30°+....+sin360°
= sin(360-350) + sin(360-340) +.......sin(360-190)+ sin(180) +sin(190)+.....sin(350)+sin(360)
=-sin(350) + -sin(340) +......-sin(190)+sin(180)+ sin(190) +.....sin(350) +sin(360)
Here all the terms will get cancelled except sin(180) and sin (360)

= [-sin(350)+sin(350)] + [-sin(340)+ sin(340)]+.....[-sin(190)+sin(190)] + sin(180)+sin(360)
=0+0+.......sin(180)+ sin 360

BUT sin 360=sin 180 =0
Therefore sin10°+sin20°+sin30°+....+sin360°=0

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