Write the value of sin10+sin20+sin30+....+sin360.Show me the solution plz

= sin(10+36x5)sin(37x5)/ sin5

= -sin10.-sin5/sin5

= sin10

eq 1 is a formula

sin(a)+sin(a+b)+sin(a+2b)............sin(a+(n-1)b) = sin(a+(n-1)b/2)sin(nb/2)/ sin(b/2)

sin10+sin20+sin30+....+sin360=0 sin10+sin20+sin30+sin40+...sin180+sin(360-170)+......+sin(360-40)+sin(360-30)+sin(360-20)+sin360-10)+sin360 sin360-x=-sinx hence all terms cancel and  sin10+sin20+sin30+....+sin360=0

a = 10°, b = 10°, then the value of n is a+(n-1)b = 360°=> 10° + (n-1) × 10° = 360°=> (n-1) × 10° = 360 - 10°=> (n-1) × 10° = 350°=> (n-1) = 350°/10°=> (n-1) = 35°=> n = 36°Sin(360°/2).Sin{(36×10)/2}/Sin(10/2) = Sin(180°).Sin(180°)/Sin(5°) = 0

Dear Student,
Please find below the solution to your problem.

sin10+sin20+sin30+....+sin360=0
sin10+sin20+sin30+sin40+...sin180+sin(360-170)+......+sin(360-40)+sin(360-30)+sin(360-20)+sin360-10)+sin360
sin360-x=-sinx
hence all terms cancel and
sin10+sin20+sin30+....+sin360=0

Thanks and Regards