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

Question

Show that

A_{n} = \sum_{k = 1}^{n} \sin k

is bounded?

Answer 1

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

\frac{\cos (0.5) - \cos (n + 0.5)}{2 \sin (0.5)}

We can now obtain the desired bound for

| A_{n} |

. 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) = 2 \sin (\frac{n}{2}) \sin (\frac{n}{2} + 0.5)

Answer 2

Hint: Since

\sin x = \frac{e^{i x} - e^{- i x}}{2 i}

we can rewrite

\sum_{n = 1}^{K} \sin n = \frac{1}{2 i} \sum_{n = 1}^{K} {(e^{i})}^{n} - \frac{1}{2 i} \sum_{n = 1}^{K} {(e^{- i})}^{n}

and both of these are geometric series.

Answer 3

A more general formula would be:

A_{n} = \sum_{k = 1}^{n} \sin k θ

= \frac{\sin θ + \sin n θ - \sin (n + 1) θ}{2 (1 - \cos θ)}

So

A_{n}

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 = 2 \sin (\frac{α + β}{2}) \cos (\frac{α - β}{2})

.

Expert Answer