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: <em>sum summatory trigonometric trig function sine sin trigonometry</em>
Yes your answer ia correct
Answer:
x = 2
Step-by-step explanation:
The solution set can be found by dividing the inequality by the coefficient of x.
6x > 7
x > 7/6
The smallest integer greater than 7/6 is 12/6 = 2.
The smallest integer solution is x = 2.
<span>Critical points are where the derivative is 0, i.e. where it crosses the x - axis
The Critical points lies where the derivative is 0, while it crosses the x-axis, SO, in this case the choice 3 looks like best answer for this.
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