AoPS - Problem

Problem

Source: USAMO 1996

Tags: trigonometry, ratio, complex numbers, geometric series, algebra

MithsApprentice

23.10.2005 02:43

Prove that the average of the numbers $n \sin n^{\circ} \; (n = 2,4,6,\ldots,180)$ is $\cot 1^{\circ}$ .

SamE

18.03.2006 08:23

Assume all angles are in degrees. Recall the basic trig identities

$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$ ,

$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$ ,

$\sin(2\theta)=2\sin\theta\cos\theta$ ,

$\cos(2\theta)=2\cos^2\theta-1$ ,

$\sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin\alpha\cos\beta$ ,

$\cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos\alpha\cos\beta$ ,

$\sin\theta=\cos(90-\theta)$ and

$\sin\theta=\sin(180-\theta)$ . Let

$x$ equal the desired average, that is

$\begin{eqnarray*}x&=&\frac{2\sin2+4\sin4+\cdots+180\sin180}{90}\\&=&\frac1{90}(2\sin2+4\sin4+\cdots+178\sin178)\\&=&\frac1{90}(2\sin2+4\sin4+\cdots+88\sin88+90\sin90+92\sin88+\cdots+178\sin2)\\&=&\frac1{90}(180\sin2+180\sin4+\cdots+180\sin88+90\sin90)\\&=&2(\sin2+\sin4+\cdots+\sin88)+1\\&=&2(\cos2+\cos4+\cdots+\cos88)+\cos0+\cos90\\&=&(\cos0+\cos2)+(\cos2+\cos4)+(\cos4+\cos6)+\cdots+(\cos86+\cos88)+(\cos88+\cos90)\\&=&2\cos1\cos1+2\cos3\cos1+2\cos5\cos1+\cdots+2\cos87\cos1+2\cos89\cos1\\&=&2\cos1(\cos1+\cos3+\cdots+\cos87+\cos89)\\&=&2\cot1\sin1(\cos1+\cos3+\cdots+\cos89)\\&=&\cot1(2\sin1\cos1+2\sin1\cos3+\cdots+2\sin1\cos89)\\&=&\cot1(\sin2-\sin0+\sin4-\sin2+\cdots+\sin90-\sin88)\\&=&\cot1(\sin90-\sin0)\\&=&\cot1,\end{eqnarray*}$ as desired.

4everwise

31.12.2006 21:35

Instead of proving that

$\frac{2\sin 2^\circ+4\sin 4^\circ+6\sin 6^\circ+\cdots+180\sin 180^\circ}{90}=\cot1^\circ,$ we'll prove that

$2\sin 1^\circ\sin 2^\circ+4\sin 1^\circ\sin 4^\circ+6\sin 1^\circ\sin 6^\circ+\cdots+180\sin 1^\circ \sin 180^\circ=90\cos 1^\circ.$ By the Product to Sum formulas, our sum simplifies to

$\begin{eqnarray*}\sum_{n=1}^{90}2n \sin 1^\circ \sin 2n^\circ &=& \sum_{n=1}^{90}n[\cos(2n-1)^\circ-\cos(2n+1)^\circ] \\ &=&-90\cos 181^\circ+\sum_{n=1}^{90}\cos(2n-1)^\circ .\\ &=& 90\cos 1^\circ+\sum_{n=1}^{90}\cos(2n-1)^\circ . \end{eqnarray*}$ It now suffices to show that

$\sum_{n=1}^{90}\cos(2n-1)^\circ =0.$ Using the identity

$\cos n^\circ =-\cos (180-n)^\circ,$

$\begin{eqnarray*}\sum_{n=1}^{90}\cos(2n-1)^\circ &=& \cos 1^\circ+\cos 3^\circ++\cdots+\cos 89^\circ+\cos 91^\circ+\cdots+\cos 177^\circ+179^\circ \\ &=& \cos 1^\circ+\cos 3^\circ+\cdots+\cos 89^\circ-(\cos 89^\circ+\cdots+\cos 3^\circ+\cos 1^\circ)\\ &=& 0. \end{eqnarray*}$

$\mathbb{QED}$

13375P34K43V312

15.04.2007 21:54

For now, disregard

$n=180$ , and consider the sum

$\sum_{k=1}^{89}2k\sin{2k^{\circ}}$ . By the fact that

$\sin{\theta}=\sin{(180^{\circ}-\theta)}$ , we have that the first sum is equal to

$\sum_{k=1}^{89}(180-2k)\sin{2k^{\circ}}$ . Now, we add these two sums together and divide by two, to find that they are equal to

$90\sum_{k=1}^{89}\sin{2k^{\circ}}$ .

$\sin{180^{\circ}}=0$ , when we add that to the sum, the sum stays the same. Since we are taking the average of 90 terms, all that is left to show is that

${\sum_{k=1}^{90}\sin{2k^\circ}}=\cot{1^{\circ}}$ . Note that this is equal to

$\Im({\sum_{k=1}^{90}\text{cis }2k^{\circ})}$ , which we can express, by Euler's formula, as

$\Im(\sum_{k=1}^{90}e^{2k\pi i/180})$ . This is the imaginary part of a finite geometric series with first term

$e^{2k\pi i/180}$ , common ratio

$e^{2k\pi i/180}$ , and 90 terms. By the formula for the sum of a geometric series, we have

$\Im(\sum_{k=1}^{90}e^{2k\pi i/180})$

$=\Im(\frac{e^{2\pi i/180}(e^{\pi i}-1)}{e^{2\pi i/180}})$

$=\Im(\frac{-2e^{\pi i/90}}{e^{\pi i/90}-1})$

$=\Im(\frac{-2e^{\pi i/90}(e^{-\pi i/90-1})}{(e^{\pi i/90}-1)(e^{\pi i/90}-1)})$

$=\Im(\frac{-2+2e^{\pi i/90}}{2-e^{-\pi i/90}-e^{\pi i/90}})$

$=\Im(\frac{-2+2\cos{2^{\circ}}+2i\sin{2^{\circ}}}{2-\cos{(-2^{\circ})}-i\sin{(-2^{\circ})}-\cos{2^{\circ}}-%Error. "isin" is a bad command. {(-2^{\circ})}})$

$=\Im(\frac{-2+2\cos{2^{\circ}}+2i\sin{2^{\circ}}}{2-2\cos^{2^{\circ}}})$

$=\frac{\sin{2^{\circ}}}{1-\cos{2^{\circ}}}$ The cotangent half angle formula tells us that

$\cot{\frac{\theta}{2}}=\frac{\sin{\theta}}{1-\cos{\theta}}$ . Applying this, we find that

$\cot{\frac{2^{\circ}}{2}}$

$=\cot{1^{\circ}}$ , as desired.

$\Box$

$\mathbb{QED}$

boxedexe

15.04.2007 22:10

This type of problem is trivial once you know about complex numbers; I believe that they should not be considered mathematical Olympiad problems anymore.

jatin

21.02.2012 16:22

Well, this problem is flattened by the following formula: Let $a_1,a_2,\cdots ,a_n$ be an A.P. with common difference $d$ . Then $\sin a_1+\sin a_2+\cdots +\sin a_n= \frac{\sin(\frac{a_1+a_n}{2})\sin (\frac{nd}{2})}{\sin(\frac{d}{2})}$ . As a sidenote, we also have $\cos a_1+\cos a_2+\cdots +\cos a_n= \frac{\cos(\frac{a_1+a_n}{2})\sin (\frac{nd}{2})}{\sin(\frac{d}{2})}$ .

tc1729

13.05.2012 23:08

We have $\frac{2\sin2+4\sin4+\cdots+180\sin180}{90}=\frac{1}{90}(2\sin2+4\sin4+\cdots+178\sin178)$ Now, we use $\sin\theta = \sin (180-\theta)$ to reduce this to $\begin{align*}\frac{1}{90}\left(\sum_{j=1}^{44}2j\sin2j+\sum_{j=0}^{44}(90+2j)\sin(90-2j)\right) &=\frac{1}{90}\left[\left(\sum_{j=1}^{44}180\sin(2j)\right) + 90\sin90\right] \\ &=2\left(\sum_{j=1}^{44}\sin 2j\right)+1\\ &=2\left(\sum_{j=1}^{44}\cos 2j\right)+\cos0+\cos90\\ &= \sum_{j=1}^{45}(\cos (2j-2)+\cos (2j))\\ \end{align*}$ By sum to product, this is $2\sum_{j=0}^{44} \cos 1\cos (2j+1)$ Factor out $\cos 1$ $2\cos1\left(\sum_{j=0}^{44}\cos (2j+1)\right) = 2\cot1\sin1\left(\sum_{j=0}^{44}\cos (2j+1)\right)$ Distribute the $2\sin 1$ $\begin{align*} \cot1\left(\sum_{j=0}^{44}2\sin1\cos(2j+1)\right) &= \cot1\left(\sum_{j=1}^{45}(\sin (2j) - \sin (2j-2))\right) \\ &=\cot1(\sin90-\sin0)\\ &=\cot1\\ \end{align*}$ as desired. $\Box$

sayantanchakraborty

09.05.2014 10:31

My proof is nice and simple: $2\sin2\sin1+2(2\sin4\sin1)+3(2\sin6\sin1)+....+90(2\sin180\sin1)$ $=\cos1-\cos3+2(\cos3-\cos5)+3(\cos5-\cos7)+....+90(\cos179-\cos181)$ $=\cos1+\cos3+\cos5+...+\cos179-90\cos181$ $=(\cos1+\cos179)+(\cos3+\cos177)+...+(\cos89+\cos91)+90\cos1$ $90\cos1$ . Dividing the first and the last line by $90\sin1$ the result follows.... Note:All the angles are in degrees.

va2010

31.01.2015 19:11

Sorry to revive, but I don't think I'm just restating anything here.

We apply vincenthuang bashing Recall that if

$\omega = cis(\theta)$ ,

$|\omega| = 1$ . Now observe that

$\frac{\omega-\omega^{-1}}{2i} = \frac{\omega - \frac{|\omega|^2}{\omega}}{2i} = \frac{2i\sin{\theta}}{2i} = \sin{\theta}$ Now observe that the desired sum is

$2\sin{2} + 2\sin{4} \cdots +2\sin{88}+1$ Let

$\omega = cis(2)$ and observe that

$\cot{1} = \frac{i(\omega^{\frac{1}{2}} + \omega^{\frac{-1}{2}})}{\omega^{\frac{1}{2}} - \omega^{\frac{-1}{2}}} = \frac{i(\omega+1)}{\omega-1}$ The desired sum is

$2\sin{2} + 2\sin{4} \cdots +2\sin{88}+1$

$= \frac{\omega-\omega^{-1}}{i} + \frac{\omega^2-\omega^{-2}}{i} \cdots +\frac{\omega^{44}-\omega^{-44}}{i}+1$

$= \frac{(\omega^{44}+\omega^{43}\cdots + \omega)-(\omega^{-44} + \omega^{-43} \cdots + \omega^{-1})}{i} +1$

$= \frac{(\omega^{45}-1)\cdot(\frac{1}{\omega} + \frac{1}{\omega^2} \cdots +\frac{1}{\omega^{44}})}{i} +1$

$=\frac{(i-1)(\frac{\omega}{\frac{\omega^{44}-1}{\omega-1}})}{i} + 1$

$=\frac{(i-1)(i-\omega)}{(\omega-1)(i)(\omega^{45})}+1$

$=(1-i)(\frac{i-\omega}{\omega-1}) + 1$

$= \frac{(1-i)(i-\omega)+\omega-1}{\omega-1}$

$= \frac{i+1-\omega+i\omega+\omega-1}{\omega-1}$

$= \frac{i\omega+i}{\omega-1}$

$= \frac{i(\omega+1)}{\omega-1}$

$= \cot{1}$ , as desired.

tauros

23.05.2020 07:42

im pretty bad at solving problems and my solutions are like super long. also im bad at proof writing too

Let

$S$ be the sum of the numbers

$n \sin n^{\circ} \; (n = 2,4,6,\ldots,180).$ In summation notation,

$\boxed{S = \sum_{n=1}^{90} 2n \sin (2n)^\circ} .$ Note: Although

$\sin 180^\circ$ is

$0$ , the problem asks for the average so it is important to keep in mind that

$S$ has 90 terms. Let

$T$ be the sum of the numbers

$n \sin (180 - n)^{\circ} \; (n = 2,4,6,\ldots,178).$ In summation form,

$T = \boxed{ \sum_{n=1}^{89} 2n \sin (180-2n)^\circ} .$ We want to show that the values of

$S$ and

$T$ are equal. Because

$\sin \theta = \sin (180 - \theta)$ for any

$\theta ,$ we can change

$T$ from

$\sum_{n=1}^{89} 2n \sin (180-2n)^\circ$ to

$\sum_{n=1}^{89} 2n \sin (2n)^\circ .$ Without the last term, the value of the two sums would be equal, but

$180\sin 180^\circ = 0$ , so

$S = T.$ Now notice what happens when we add

$S$ to

$T.$ We know

$S+T$ is just

$2S$ ; this will be the LHS. However, notice that for every term

$2n \sin (2n)^\circ$ in

$S$ , there is a corresponding term

$(180 - 2n)\sin (2n)^\circ$ in

$T$ , and these terms sum to

$180\sin (2n)^\circ .$ So,

$S + T = 2S = \sum_{n=1}^{90} 180 \sin (2n)^\circ$ (don’t forget the

$180\sin 180^\circ$ !) The original average is

$\dfrac{S}{90} = \dfrac{2S}{180} = \sum_{n=1}^{90} \sin (2n)^\circ .$ We want to show that this is equal to

$\cot 1^\circ .$ Since I am not smart enough, I had to prove it with complex numbers (used radians instead because that is the proper notation when using exponential form). Because

$e^{i \theta} = \cos \theta + i\sin \theta ,$ we can change

$\dfrac{S}{90}$ into

$\sum_{n=1}^{90} \text{Im} \left( e^{2n\pi i / 180} \right) = \text{Im} \left( \sum_{n=1}^{90} e^{2n\pi i / 180} \right) .$ The sum we are taking the imaginary part of is a 90-term geometric series with first term

$e^{2\pi i / 180} = e^{\pi i / 90}$ and common ratio

$e^{\pi i / 90} .$ So, using the geometric sum formula, we have

$\begin{align*} \text{Im} \left( \dfrac{e^{\pi i /90} (e^{90\pi i /90} - 1)}{e^{\pi i /90} - 1} \right) \\ \\ =\text{Im} \left( \dfrac{-2 e^{\pi i /90}}{e^{\pi i /90} - 1} \right) \\ \\ =\text{Im} \left( \dfrac{-2 \cos \tfrac{\pi}{90} - 2i\sin \tfrac{\pi}{90}}{\cos \tfrac{\pi}{90} - 1 + i\sin \tfrac{\pi}{90}} \cdot \dfrac{\cos \tfrac{\pi}{90} - 1 - i\sin \tfrac{\pi}{90}}{\cos \tfrac{\pi}{90} - 1 - i\sin \tfrac{\pi}{90}} \right) \\ \\ =\text{Im} \left( \dfrac{-2\cos^2 \tfrac{\pi}{90} +2\cos \tfrac{\pi}{90} + 2i\cos \tfrac{\pi}{90}\sin \tfrac{\pi}{90} - 2i\cos \tfrac{\pi}{90}\sin \tfrac{\pi}{90} + 2i\sin \tfrac{\pi}{90} - 2\sin^2 \tfrac{\pi}{90} }{(\cos \tfrac{\pi}{90} - 1)^2 + \sin^2 \tfrac{\pi}{90}} \right) \\ \\ =\dfrac{2\sin \tfrac{\pi}{90}}{\cos^2 \tfrac{\pi}{90} - 2\cos \tfrac{\pi}{90} + 1 + \cos^2 \tfrac{\pi}{90}} = \dfrac{2\sin \tfrac{\pi}{90}}{2 - 2\cos \tfrac{\pi}{90}} = \dfrac{2\sin \tfrac{\pi}{90}}{2 - 2\cos \tfrac{\pi}{90}} = \dfrac{1\sin \tfrac{\pi}{90}}{1 - \cos \tfrac{\pi}{90}} \\ \\ =\dfrac{1}{\dfrac{ 1 - \cos \tfrac{\pi}{90} }{1\sin \tfrac{\pi}{90}}} = \dfrac{1}{\tan \dfrac{ \tfrac{\pi}{90}}{2}} = \cot \tfrac{\pi}{180} = \cot 1^\circ . \blacksquare . \end{align*}$

AopsUser101

25.05.2020 02:53

$\sum_{n = 1}^{90} (2n) \sin (2n) = \left( \sum_{n = 1}^{44} 180 \sin (2n) \right) + 90 \sin 90$ , so the average of the numbers is $\left( \sum_{n = 1}^{44} 2 \sin (2n) \right) + 1$ . Note that $\sum_{n = 1}^{44} 2 \sin 1 \sin (2n) = \sum_{n = 1}^{44} \cos (2n-1) - \cos (2n + 1)$ , which clearly telescopes to $-\cos (89) + \cos (1)$ . Thus, $\sum_{n = 1}^{44} \sin (2n) = \frac{-\cos (89) + \cos (1)}{\sin (1)} = \frac{-\sin 1 + \cos 1}{\sin 1} = -1 + \cot 1$ The problem statement follows.

PROA200

28.06.2020 20:59

Nothing too new here. We use "Vincent Huang bashing."

Let

$z=e^{\frac{i\pi}{180}}$ . Then

$\begin{align*} &\sum_{k=1}^{90} 2k(\sin{2k^{\circ}})\\ =&\sum_{k=1}^{90} 2k\cdot \frac{1}{2i}\left(z^{2k}-\frac{1}{z^{2k}}\right)\\ =&\frac{1}{i}\sum_{k=1}^{90}k\left(z^{2k}-\frac{1}{z^{2k}}\right)\\ =&\frac{1}{i}\left[\left(z^2+2z^4+3z^6+\dots+90z^{180}\right)-\left(\frac{1}{z^2}+\frac{2}{z^4}+\dots+\frac{90}{z^{180}}\right)\right] \end{align*}$ The key observation here is that

$z^{180}=-1$ . So we multiply everything on the right-hand section by

$-z^{180}$ . Then it becomes

$\begin{align*} &\frac{1}{i}\left[\left(z^2+2z^4+3z^6+\dots+90z^{180}\right)-\left(\frac{1}{z^2}+\frac{2}{z^4}+\dots+\frac{90}{z^{180}}\right)\right]\\ =&\frac{1}{i}\left[\left(z^2+2z^4+3z^6+\dots+90z^{180}\right)+\left(z^{178}+2z^{176}+\dots+90\right)\right]\\ =&\frac{1}{i}\left(90z^2+90z^4+\dots+90z^{178}+90z^{180}+90\right) \end{align*}$ But again

$z^{180}=-1$ , so

$90$ and

$90z^{180}$ cancel. This is a geometric series, and equals

$\begin{align*} \frac{90}{i}\left(\frac{z^{180}-1}{z^2-1}-1\right)=\frac{90}{i}\left(\frac{z^{180}-z^2}{z^2-1}\right)=90i\left(\frac{z^2+1}{z^2-1}\right)=90\cdot \frac{\cos 1^{\circ}}{\sin 1^{\circ}}, \end{align*}$ as desired.

$\blacksquare$

brainiacmaniac31

29.06.2020 02:10

(Omitting the degree symbol for ease).

$\frac{1}{90}\sum_{i=1}^{90}2i\sin 2i=\frac{1}{90}(2\sin 2+178\sin 178+4\sin 4+176\sin 176+\cdots+90\sin 90)$

$=\frac{1}{90}\left(180(\sin 2+\sin 4+\cdots+\sin 88)+90\right)=1+2\sum_{i=1}^{44}\sin 2i$

$=1+2\sum_{i=1}^{22}2\sin 45\cos(2i-1)=1+2\sqrt2\sum_{i=1}^{22}\cos(2i-1)\quad\because\text{Sum to Product Identities}.$ Now, since

$\cot 1=\frac{\cos 1}{\sin 1}$ , we can multiply through by

$\sin 1$ and show that the resulting expression is equal to

$\cos 1$ .

$\sin 1\left(1+2\sqrt2\sum_{i=1}^{22}\cos(2i-1)\right)=\sin 1+2\sqrt2(\cos1\sin1+\cos3\sin1+\cdots+\cos43\sin1)$

$=\sin1+\sqrt2(\sin2+\sin0+\sin4-\sin2+\cdots+\sin44-\sin42)\quad\because\text{Product to Sum Identities}$

$=\sin 1+\sqrt2\sin 44=\sin 1+2\sin 44\cos 45$

$=\sin 1+\sin 89-\sin 1=\sin 89=\cos 1\quad\because\text{Product to Sum Identities}.$

Keith50

10.03.2021 06:18

$\dagger \ \color{green}{\textrm{\textbf{A different solution using Abel's Summation by Parts:}}}$ As we know that $\color{blue}{\sin (a)+\sin (a+d)+\sin (a+2d)+\ldots +\sin (a+(n-1)d)=\frac{\sin \left(a+\frac{n-1}{2}d\right)\cdot \sin \left(\frac{n}{2}d\right)}{\sin \left(\frac{d}{2}\right)}}$ and $\color{magenta}{\cos (a)+\cos (a+d)+\cos (a+2d)+\ldots +\cos (a+(n-1)d)=\frac{\cos \left(a+\frac{n-1}{2}d\right)\cdot \sin \left(\frac{n}{2}d\right)}{\sin \left(\frac{d}{2}\right)}}$ , we apply Abel's Summation by Parts Formula to get: $\begin{align*} &\frac{1}{90}\sum_{n=1}^{90}2n\sin (2n) \\ =&\frac{1}{90}\left(180\cdot \frac{\sin (91)}{\sin (1)}-\frac{2}{\sin (1)}\sum_{n=1}^{89} (\sin(n+1)\cdot \sin (n))\right) \\ =&2\cot (1)-\frac{1}{90\sin (1)} \sum_{n=1}^{89} (\cos (1) - \cos (2n+1)) \\=& 2\cot (1)-\cot (1)+\frac{2}{\sin (1)}\sum_{n=0}^{89} \cos (2n+1) \\ =& \cot (1)+\frac{2}{\sin (1)} \cdot \frac{\cos (1+89)\cdot \sin (90)}{\sin (1)}\\ =&\cot (1). \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \blacksquare \end{align*}$

v_Enhance

01.05.2021 06:09

Solution from Twitch Solves ISL: Because $n \sin n^{\circ} + (180-n) \sin(180^{\circ}-n^{\circ}) = 180 \sin n^{\circ}$ So enough to show that $\sum_{n=0}^{89} \sin (2n)^{\circ} = \cot1^{\circ}$ Let $\zeta = \cos2^{\circ} + i \sin 2 ^{\circ}$ be a primitive root. Then $\begin{align*} \sum_{n=0}^{89} \frac{\zeta^n - \zeta^{-n}}{2i} &= \frac{1}{2i} \left[ \frac{\zeta^{90}-1}{\zeta-1} - \frac{\zeta^{-90} - 1}{\zeta^{-1}-1} \right] \\ &= \frac{1}{2i} \left[ \frac{-2}{\zeta-1} - \frac{-2}{\zeta^{-1}-1} \right] \\ &= \frac{1}{-i} \frac{\zeta^{-1}-\zeta}{(\zeta-1)(\zeta^{-1}-1)} = i \cdot \frac{\zeta+1}{\zeta-1}. \end{align*}$ Also, $\begin{align*} \cot 1^{\circ} &= \frac{\cos 1^{\circ}}{\sin 1^{\circ}} = \frac{(\cos1^{\circ})^2}{\cos1^{\circ}\sin1^{\circ}} \\ &= \frac{\frac{\cos2^{\circ}+1}{2}}{\frac{\sin2^{\circ}}{2}} = \frac{\frac{1}{2}(\zeta+\zeta^{-1})+1}{\frac{1}{2i}(\zeta-\zeta^{-1})} \\ &= i \cdot \frac{(\zeta+1)^2}{\zeta^2-1} = i \cdot \frac{\zeta+1}{\zeta-1}. \end{align*}$ So we're done.

vsamc

25.06.2021 06:01

not a desirable method oops Work in radians. Let $S$ be the set of even numbers between $2$ and $180$ inclusive. Then, it suffices to show that $\sum_{n\in S} n\sin \frac{n\pi}{180} = 90 \cot \frac{\pi}{180}$ . However, since $n \sin n = \Im(ne^{n\pi i / 180})$ , it suffices to show that $\sum_{n\in S} \Im(ne^{n\pi i / 180}) = \Im(\sum_{n\in S}ne^{n\pi i / 180}) = 90 \cot \frac{\pi}{180}$ . Now, let $A = \sum_{n\in S}ne^{n\pi i / 180}$ . We can write out $A$ and $e^{2\pi i / 180}A$ out and subtract to get $(1 - e^{2\pi i / 180})A = 2(e^{2i\pi/180} + e^{4\pi i / 180} + \cdots + e^{180\pi i / 180}) + 180e^{2\pi i / 180}.$ Now, we compute $e^{2i\pi/180} + e^{4\pi i / 180} + \cdots + e^{180\pi i / 180}$ . Let $z = e^{2\pi i / 180}$ ; then this sum becomes $z + z^2 + \cdots + z^{90} = z(1 + z +\cdots + z^{89}) = z\left(\frac{1 - z^{90}}{1-z}\right) = z\left(\frac{2}{1-z}\right) = \frac{2z}{1-z} = \frac{2z\left(1 - \frac{1}{z}\right)}{(1-z)\left(1 - \frac{1}{z}\right)} = \frac{2z - 2}{2 - 2\Re(z)} = -\frac{1-z}{1 - \Re(z)}.$ Thus, we can write $(1-z)A = 2\left(-\frac{1-z}{1-\Re(z)}\right) + 180z \iff A = \frac{-\frac{2-2z}{1-\Re(z)} + 180z}{1-z} = -\frac{2}{1-\Re(z)} + \frac{180z}{1-z}.$ We want to find $\Im(A)$ . However, since $-\frac{2}{1-\Re(z)}$ is real, it suffices to find the imaginary part of $\frac{180z}{1-z} = \frac{180z\left(1 - \frac{1}{z}\right)}{(1-z)\left(1 - \frac{1}{z}\right)} = \frac{180z - 180}{2 - 2\Re(z)}.$ Since $-\frac{180}{2-2\Re(z)}$ is real, it thus suffices to find $\Im\left(\frac{180z}{2-2\Re(z)}\right) = \Im\left(\frac{90z}{1 - \Re(z)}\right) = \frac{90\Im(z)}{1 - \Re(z)} = \frac{90 \sin \frac{2\pi}{180}}{1 - \cos \frac{2\pi}{180}}.$ Now, using double angle, this is equal to $\frac{180\sin \frac{\pi}{180}\cos \frac{\pi}{180}}{1 - (1 - 2\sin^2 \frac{pi}{180})} = \frac{180\sin \frac{\pi}{180}\cos \frac{\pi}{180}}{2\sin^2 \frac{\pi}{180}} = \frac{90\cos \frac{\pi}{180}}{\sin \frac{\pi}{180}} = 90 \cot \frac{\pi}{180},$ as desired.

john0512

26.12.2022 11:07

We rewrite what we want to prove as $\sum_{n=1}^{90} n \sin 2n = 45\cot 1.$ Note that $n\sin(2n)+(90-n)\sin(2(90-n))=90\sin(2n),$ so this becomes $45+90\sum_{n=1}^{44}\sin(2n)=45\cot(1),$ where the 45 comes from the isolated term $45\sin(90).$ We can divide out the 45 to get $1+2\sum_{n=1}^{44}\sin(2n)=\cot(1).$ Multiplying this by $\sin(1),$ this becomes $\sum_{n=1}^{44}2\sin(2n)\sin(1)=\cos(1)-\sin(1).$ Then, by product-to-sum, this becomes $\sum_{n=1}^{44}\cos(2n-1)-\cos(2n+1)=\cos(1)-\sin(1).$ This is true since we now have a happy tale of telescoping: $LHS=\cos(1)-\cos(3)+\cos(3)-\cos(5)+\cos(5)\cdots -\cos(89)=\cos(1)-\cos(89)=\cos(1)-\sin(1),$ so we are done.

HamstPan38825

09.01.2023 18:32

First, notice that $n \sin n^\circ + (180-n) \sin(180-n)^\circ = 180 \sin n^\circ,$ so the condition is equivalent to $\sum_{n=0}^{89} \sin(2n)^\circ = \cot 1^\circ.$ This is just a complex bash: note that $\sum_{n=0}^{89} \frac{\omega^n-\omega^{-n}}{2i} = \frac{\frac 1{1-\omega} - \frac 1{1-\omega^{-1}}}{i} = \frac{\omega-\omega^{-1}}{i(1-\omega^{-1})(1-\omega)}$ where $\omega = \exp\left(\frac{\pi}{90}\right)$ . However, for $\zeta = \exp\left(\frac{\pi}{180}\right)$ , we have $\cot 1^\circ = \frac{\zeta+\zeta^{-1}}{-i(\zeta-\zeta^{-1})} = \frac{\zeta^2- \zeta^{-2}}{i(1-\zeta^{-2})(1-\zeta^2)},$ as required.