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
