Show that An=∑k=1nsin⁡k is bounded?

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Addisyn Thompson · Accepted Answer

Note that  2(sin⁡k)(sin⁡(0.5))=cos⁡(k−0.5)−cos⁡(k+0.5)  This is obtained by using the ordinary expression for the cosine of a sum.  Add up, k=1 to n. On the right, there is mass cancellation. We get  cos⁡(0.5)−cos⁡(n+0.5)  Thus our sum of sines is  cos⁡(0.5)−cos⁡(n+0.5)2sin⁡(0.5)  We can now obtain the desired bound for |An|. For example, 2 works, but not by much.  We could modify the appearance of the above formula by using the fact that cos⁡(0.5)−cos⁡(n+0.5)=2sin⁡(n2)sin⁡(n2+0.5)

dikgetse3u · Accepted Answer

Hint: Since  sin⁡x=eix−e−ix2i  we can rewrite  ∑n=1Ksin⁡n=12i∑n=1K(ei)n−12i∑n=1K(e−i)n  and both of these are geometric series.

RizerMix · Accepted Answer

A more general formula would be: An=∑k=1nsin⁡kθ =sin⁡θ+sin⁡nθ−sin⁡(n+1)θ2(1−cos⁡θ) So An is clearly bounded (just simply check the case where θ=1). The formula can be proved by induction using the trig identity: sin⁡α+sin⁡β=2sin⁡(α+β2)cos⁡(α−β2).

Show that A_n=\sum_{k=1}^n\sin k is bounded?

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