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1 month ago

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

1 answer:

4 0

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

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1 month ago

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14 days ago

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Response:

6 hours worked as a tutor

Detailed clarification:

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10 days ago

Before the last school year began, it was estimated that the average discretionary personal expenses each school year for a stud

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Answer:

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Given the estimate of $2,110, the average is indeed higher than this estimate.

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26 days ago

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1 month ago