Sin 1 + sin 2 + sin3 .................+ sin 360 = ?

Assuming arcs are measured in degrees, let S represent the following sum:

S = sin 1° + sin 2° + sin 3° +... + sin 359° + sin 360°


By rearranging, S can be reformulated as

S = [sin 1° + sin 359°] + [sin 2° + sin 358°] +... + [sin 179° + sin 181°] + sin 180° +
+ sin 360°

S = [sin 1° + sin(360° – 1°)] + [sin 2° + sin(360° – 2°)] +... + [sin 179° + sin(360° – 179)°]
+ sin 180° + sin 360°          (i)


However, for any real k,

sin(360° – k) = – sin k


Thus,

S = [sin 1° – sin 1°] + [sin 2° – sin 2°] +... + [sin 179° – sin 179°] + sin 180° + sin 360°

S results in 0 + 0 +... + 0 + 0 + 0        (... since sine of 180° and 360° are both equal to 0)

Therefore, S equals 0.


Each pair within the brackets negates itself, leading the sum to total zero.

∴   sin 1° + sin 2° + sin 3° +... + sin 359° + sin 360° equals 0.          ✔


I hope this clarifies things. =)


Tags:  sum summatory trigonometric trig function sine sin trigonometry