Question

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

Answer 1

Assuming that arcs are given in degrees, call S the following sum:

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


Rearranging the terms, you can rewrite S 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)


But for any real k,

sin(360° – k) = – sin k


then,

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

S = 0 + 0 + ... + 0 + 0 + 0        (... as sin 180° = sin 360° = 0)

S = 0


Each pair of terms in brackets cancel out themselves, so the sum equals zero.

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


I hope this helps. =)


Tags:  sum summatory trigonometric trig function sine sin trigonometry