We need \(S=\displaystyle\sum_{k=1}^{90}2k\sin(2k^\circ)\), and the average is \(S/90\).

Use \(2\sin(2k^\circ)\sin1^\circ=\cos((2k-1)^\circ)-\cos((2k+1)^\circ)\), so \(\sin(2k^\circ)=\dfrac{\cos((2k-1)^\circ)-\cos((2k+1)^\circ)}{2\sin1^\circ}\).

Then \(S=\dfrac{1}{\sin1^\circ}\displaystyle\sum_{k=1}^{90}k\left[\cos((2k-1)^\circ)-\cos((2k+1)^\circ)\right]\). Writing \(c_m=\cos(m^\circ)\), by an Abel-summation (telescoping) rearrangement:

\(\displaystyle\sum_{k=1}^{90}k(c_{2k-1}-c_{2k+1})=c_1+c_3+c_5+\cdots+c_{179}-90\,c_{181}\)

Now \(c_1+c_3+\cdots+c_{179}\) (the 90 odd-degree cosines from \(1^\circ\) to \(179^\circ\)) pair up as \(\cos\theta+\cos(180^\circ-\theta)=\cos\theta-\cos\theta=0\) for each pair \((\theta,180^\circ-\theta)\), so this sum is 0.

Also \(c_{181}=\cos181^\circ=-\cos1^\circ\), so \(-90c_{181}=90\cos1^\circ\).

Hence \(\displaystyle\sum_{k=1}^{90}k(c_{2k-1}-c_{2k+1})=0+90\cos1^\circ=90\cos1^\circ\).

Therefore \(S=\dfrac{90\cos1^\circ}{\sin1^\circ}=90\cot1^\circ\).

Average \(=\dfrac{S}{90}=\cot1^\circ\).