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: $$\begin{eqnarray*}2\Im\left(\frac{92 z-90 z^{2}}{(1-z)^2}\right)&=&2\Im\left(90\frac{z}{1-z}+\frac{2z}{(1-z)^2}\right)=90\cdot\Im\left(\frac{2e^{2it}}{1-e^{2it}}\right)\\&=&90\cdot\frac{\Im(ie^{it})}{\sin t}=90\cot t.\end{eqnarray*}$$

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

  • $\begingroup$ I don't believe I've learned using e yet. Is there a simpler way? $\endgroup$ – Mathy Person Jan 12 '15 at 1:43
  • $\begingroup$ 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. $\endgroup$ – alex.jordan Jan 12 '15 at 1:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.