Another way to express $\text{cis } 75^\circ + \text{cis } 83^\circ + \text{cis } 91^\circ + \dots + \text{cis } 147^\circ$

Question

The number $$\text{cis } 75^\circ + \text{cis } 83^\circ + \text{cis } 91^\circ + \dots + \text{cis } 147^\circ$$ is expressed in the form $r \, \text{cis } \theta$, where $0 \le \theta < 360^\circ$. Find $\theta$ in degrees.

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Hint: $\text{cis}\ \theta=\cos \theta+i \sin \theta$

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Edit:

I simplified the expression down to$$\frac{\text{cis} \ 75^\circ \sin 40^\circ \ \text{cis} \ 40^\circ }{\sin 4^\circ \ \text{cis} \ 4^\circ}.$$

What should I do now?

Answer 1

HINT:

By How to prove Euler's formula: $e^{it}=\cos t +i\sin t$? cis$(t)=e^{it},$

we need $$\sum_{r=0}^9(e^i)^{(75+8r)^\circ}=\sum_{r=0}^9e^{75^\circ i} (e^{8i^\circ})^r$$ which is in Geometric progression.

See also: How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

Answer 2

Hint:

$e^{i(75\pi/180)}+e^{i(83\pi/180)}+e^{i(91\pi/180)}+\cdots +e^{i(147\pi/180)}$

G.P series with common ratio $r=e^{i\frac{8\pi}{180}}$

Sum of GP series $S_n = \frac{1-r^n}{1-r}$

Answer 3

\begin{align} &\text{cis} 75 ^\circ+\text{cis} 83 ^{\circ}+\ldots + \text{cis}147^\circ \\ &= \text{cis} 75^{\circ}(1+\text{cis}8 ^\circ+\ldots+\text{cis}72^{\circ}) \\ \end{align}

Using the geometric series formula, the answer is the same as:

The expression $\frac{(\text{cis}\ 75^\circ)(\text{cis}\ 80^\circ−1)}{\text{cis}\ 8^\circ−1}$ can be written as $r\ \text{cis}\ \theta$.