What is the value of this trigonometric sum ratio:∑n=144cos⁡n∘∑n=144sin⁡n∘=?

Note: cos⁡n∘=cos⁡π180nso the above sums can be expressed in terms of geometric series, i.e.∑n=144eiπn180=eiπ45180−eiπ1801−eiπ180=eπ4−eiπ1801−eiπ180This simplifies to:(eiπ4−eiπ180)(1−e−iπ180)4sin2π360=ei23π1802sin⁡(22π180)e−iπ360i2sin⁡(π360)4sin2π360which in turn becomes−ei45π360sin⁡(22π180)sin⁡(π360)We seek the ratio of the real part of this expression to the imaginary part. The result is∑n=144cos⁡n∘∑n=144sin⁡n∘=cot⁡(π8)=2+1

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Payton Benson

Answered question

2022-03-15

What is the value of this trigonometric sum ratio:

\frac{\sum_{n = 1}^{44} {\cos n}^{\circ}}{\sum_{n = 1}^{44} {\sin n}^{\circ}} = ?

Answer & Explanation

Curtiello69r

Beginner2022-03-16Added 6 answers

Note:

{\cos n}^{\circ} = \cos \frac{π}{180} n

so the above sums can be expressed in terms of geometric series, i.e.

\sum_{n = 1}^{44} e^{i π \frac{n}{180}} = \frac{e^{i π \frac{45}{180}} - e^{i \frac{π}{180}}}{1 - e^{i \frac{π}{180}}} = \frac{e^{\frac{π}{4}} - e^{i \frac{π}{180}}}{1 - e^{i \frac{π}{180}}}

This simplifies to:

\frac{(e^{i \frac{π}{4}} - e^{i \frac{π}{180}}) (1 - e^{- i \frac{π}{180}})}{4 \frac{\sin^{2} π}{360}}

= \frac{e^{i 23 \frac{π}{180}} 2 \sin (22 \frac{π}{180}) e^{- i \frac{π}{360}} i 2 \sin (\frac{π}{360})}{4 \sin^{2} π 360}

which in turn becomes

- e^{i 45 \frac{π}{360}} \frac{\sin (22 \frac{π}{180})}{\sin (\frac{π}{360})}

We seek the ratio of the real part of this expression to the imaginary part. The result is

\frac{\sum_{n = 1}^{44} {\cos n}^{\circ}}{\sum_{n = 1}^{44} {\sin n}^{\circ}} = \cot (\frac{π}{8}) = \sqrt{2} + 1

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