# Prove that $\frac{1}{90} \sum_{n=1}^{90} 2n \cdot \sin((2n)^\circ) = \cot (1^{\circ})$

Show that $$\frac{(2\sin(2^\circ)) + (4\sin(4^\circ))+ (6\sin(6^\circ)) + \ldots +(180\sin(180^\circ))}{90} = \cot(1^\circ).$$

I used a lot of steps, and typing it all down on here would take me an hour, but here are my last few steps up to the point where I got stuck:

$$180 (\sin(2^\circ) + \sin(4^\circ) + \sin(6^\circ) +.....+ \sin(88^\circ)) + 90$$

Using the product-to-sum formulas, I then reduced it down to

$$180(2\sin(45^\circ) [\cos(43^\circ)+\cos(41^\circ)+\cos(39^\circ)+.....\cos(3^\circ)+\cos(1^\circ)] + 90$$

Simplifying a bit more gives me:

$$\frac{180\sqrt{2} [\cos(43^\circ)+\cos(41^\circ)+\cos(39^\circ)+.....\cos(3^\circ)+\cos(1^\circ)] + 90}{90}$$ $$= 2\sqrt{2} [\cos(43^\circ)+\cos(41^\circ)+\cos(39^\circ)+.....\cos(3^\circ)+\cos(1^\circ)] + 90.$$

Now I am stuck. What can I do next to achieve the final result?

$$S=\sum_{n=1}^{90}2(91-n)\sin(2n^\circ) = \sum_{n=1}^{90}2\sum_{k=1}^{n}\sin(2k^\circ).$$ Since: $$2(\sin 1^{\circ})\sum_{k=1}^{n}\sin(2k^\circ)=\sum_{k=1}^{n}\left(\cos((2k-1)^{\circ})-\cos((2k+1)^{\circ})\right)=\cos 1^\circ-\cos((2n+1)^\circ)$$ it follows that: $$S = \frac{1}{\sin 1^\circ}\left(90\cos 1^\circ-\sum_{n=1}^{90}\cos((2n+1)^{\circ})\right),$$ but with the same trick shown in this other question we have that: $$\sum_{n=1}^{90}\cos((2n+1)^{\circ})=-2\cot 1^\circ,$$ hence $\color{red}{S=92\cot 1^{\circ}}$. On the other hand, since: $$\sum_{n=1}^{90}2\sin(2n^\circ) = 2\cot 1^{\circ}$$ always by the same trick, it follows that: $$\sum_{n=1}^{90}2n\sin(2n^\circ)=(91\cdot 2-92)\cot 1^\circ= \color{blue}{90\cot 1^{\circ}}$$ as wanted.

For a complex-analytic derivation, set $t=\frac{\pi}{180}, z=e^{2it}$ and consider that: $$\sum_{n=1}^{90}2n \sin(2nt) = \Im\sum_{n=1}^{90}2n z^n=2\Im\left(\frac{z-91z^{91}+90 z^{92}}{(1-z)^2}\right).$$ Since $z^{90}=-1$, the last expression simplifies to: $$\begin{eqnarray*}2\Im\left(\frac{92 z-90 z^{2}}{(1-z)^2}\right)&=&2\Im\left(90\frac{z}{1-z}+\frac{2z}{(1-z)^2}\right)=90\cdot\Im\left(\frac{2e^{2it}}{1-e^{2it}}\right)\\&=&90\cdot\frac{\Im(ie^{it})}{\sin t}=90\cot t.\end{eqnarray*}$$

• How did you get the formulas found on the second line? – Mathy Person Jan 12 '15 at 2:15
• @Mathy: it is always the sum-product formula $$\cos(a-b)-\cos(a+b)=2\sin(a)\sin(b)$$ leading to a telescopic sum. – Jack D'Aurizio Jan 12 '15 at 2:17
• Is there another method to finish this problem without using a long telescoping sum? – Mathy Person Jan 12 '15 at 4:25
• @MathyPerson: are you allowed to use differentiation? In such a case, consider that: $$\sum_{n=1}^{N}n \sin(nx) = -\frac{d}{dx}\sum_{n=1}^{N}\cos(n x).$$ – Jack D'Aurizio Jan 12 '15 at 12:26
• @MathyPerson: if not, in Italy we use to say: If you want to make a pie, you have to break some eggs. – Jack D'Aurizio Jan 12 '15 at 12:31

Your sum is the imaginary part of a (complex) arithme-geometric series. Sum $2n\,e^{i(2n)^{\circ}}$ instead and take the imaginary part when this is done.

• I don't believe I've learned using e yet. Is there a simpler way? – Mathy Person Jan 12 '15 at 1:43
• So you don't know yet that $\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$? It's very helpful for evaluating this. But perhaps there is a trig identity that handles it somehow. – alex.jordan Jan 12 '15 at 1:55