prove that sin10+sin20+sin30+ ......... + sin360=0?

Question

explain

Answer 1

pair up the opposite sines, for example sin(10) and sin(350) and add them:
sin(10) + sin(350) = 0
sin(20) + sin(340) = 0
...
until you only have sin(180) and sin(360) which both equal 0

PS. I assumed you were using degrees (sexagesimal). Won't work in radians (RAD) or metric (decimal) angles.

Answer 2

Remeber this
sin x + sin 2 x +...+sin nx

sin(nx/2)sin((x+nx)/2)/sinx/2

Here
x=10
n=36
So the sum is
=sin(360/2)sin(370/2)/sin5
=sin 180 sin(sin(185)/sin 5=0
As sin 180=0
So the whole thing is 0

If you know complex numbers then it's very easy.
Let
sin10+sin20+sin30+ ......... + sin360=S
Then consider another series
cos10 +cos20 +cos30+.....+cos360=C
Multiply j with S j=sqrt(-1)
Use cos a + j sin a =exp(ja)
Then C+jS= exp(j10)+exp(j20)+exp(j30)+...+exp(j360)
Apply formula for geometric series.
C+jS= exp(j10) (exp((j10)*36)-1)/(expj10-1)
exp(j10)( exp(j360)-1)/(expj10-1)= 0
Since exp(j360)=1
Therefore C+jS= exp(j10)(0)/(expj10-1)
Implies S=0
Therefore jS=0 ,C=0
or S=0 which was to be proved
and C=0

Answer 3

just remove all numeric after sine like
sin10 to sin 0

it will prove