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

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 $\cos a_1+\cos a_2+\cdots +\cos a_n= \frac{\cos(\frac{a_1+a_n}{2})\sin (\frac{nd}{2})}{\sin(\frac{d}{2})}$.

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. $\Box$

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.

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

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

$\sum_{n = 1}^{90} (2n) \sin (2n) = \left( \sum_{n = 1}^{44} 180 \sin (2n) \right) + 90 \sin 90$, 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.

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

$\dagger \ \color{green}{\textrm{\textbf{A different solution using Abel's Summation by Parts:}}}$ 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*}

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.

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.

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.

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.