Prove that the average of the numbers nsinn∘(n=2,4,6,…,180) is cot1∘.
Problem
Source: USAMO 1996
Tags: trigonometry, ratio, complex numbers, geometric series, algebra
18.03.2006 08:23
31.12.2006 21:35
15.04.2007 21:54
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.
21.02.2012 16:22
Well, this problem is flattened by the following formula: Let a1,a2,⋯,an be an A.P. with common difference d. Then sina1+sina2+⋯+sinan=sin(a1+an2)sin(nd2)sin(d2). As a sidenote, we also have cosa1+cosa2+⋯+cosan=cos(a1+an2)sin(nd2)sin(d2).
13.05.2012 23:08
We have 2sin2+4sin4+⋯+180sin18090=190(2sin2+4sin4+⋯+178sin178) Now, we use sinθ=sin(180−θ) to reduce this to 190(44∑j=12jsin2j+44∑j=0(90+2j)sin(90−2j))=190[(44∑j=1180sin(2j))+90sin90]=2(44∑j=1sin2j)+1=2(44∑j=1cos2j)+cos0+cos90=45∑j=1(cos(2j−2)+cos(2j)) By sum to product, this is 244∑j=0cos1cos(2j+1) Factor out cos1 2cos1(44∑j=0cos(2j+1))=2cot1sin1(44∑j=0cos(2j+1)) Distribute the 2sin1 cot1(44∑j=02sin1cos(2j+1))=cot1(45∑j=1(sin(2j)−sin(2j−2)))=cot1(sin90−sin0)=cot1 as desired. ◻
09.05.2014 10:31
My proof is nice and simple: 2sin2sin1+2(2sin4sin1)+3(2sin6sin1)+....+90(2sin180sin1) =cos1−cos3+2(cos3−cos5)+3(cos5−cos7)+....+90(cos179−cos181) =cos1+cos3+cos5+...+cos179−90cos181 =(cos1+cos179)+(cos3+cos177)+...+(cos89+cos91)+90cos1 90cos1. Dividing the first and the last line by 90sin1 the result follows.... Note:All the angles are in degrees.
31.01.2015 19:11
Sorry to revive, but I don't think I'm just restating anything here.
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
25.05.2020 02:53
∑90n=1(2n)sin(2n)=(∑44n=1180sin(2n))+90sin90, so the average of the numbers is (∑44n=12sin(2n))+1. Note that ∑44n=12sin1sin(2n)=∑44n=1cos(2n−1)−cos(2n+1), which clearly telescopes to −cos(89)+cos(1). Thus, 44∑n=1sin(2n)=−cos(89)+cos(1)sin(1)=−sin1+cos1sin1=−1+cot1The problem statement follows.
28.06.2020 20:59
Nothing too new here. We use "Vincent Huang bashing."
29.06.2020 02:10
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*}
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.
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.
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.
09.01.2023 18:32
First, notice thatn \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.