Posted by: Dave Richeson | September 9, 2013

2013: the year of pi

A couple days ago I saw this tweet.

Pretty cool! Let’s see why

$\arctan(2)+\arctan(0)+\arctan(1)+\arctan(3)=\pi.$

Two terms are easy to deal with:

$\arctan(0)=0$ and

$\arctan(1)=\pi/4.$

But why is $\arctan(2)+\arctan(3)=3\pi/4?$

One way to prove this is via the inverse trigonometric identity

$\displaystyle \arctan(a)+\arctan(b)=\arctan\left(\frac{a+b}{1-ab}\right)\, (\text{mod }\pi).$

In this case

$\arctan(2)+\arctan(3)=\arctan(-1)\, (\text{mod }\pi).$

Because $\arctan(2)$ and $\arctan(3)$ are between 0 and $\pi/2$,  $\arctan(a)+\arctan(b)=3\pi/4$.

I didn’t use this approach on my first attempt to solve the problem.  (To be honest, I didn’t know this identity existed before finding it online.) I used geometry and trigonometry. We are interested in the angle $\theta$ in the diagram below.

The law of cosines tells us that

$(2+3)^2=(\sqrt{5})^2+(\sqrt{10})^2-2\sqrt{5}\sqrt{10}\cos\theta.$

This implies that $\cos\theta=\frac{1}{\sqrt{2}},$ and hence $\theta=3\pi/4.$

(I wonder if there is a year of tau coming up sooner than the year 112233.)

Update: I just saw that Cut-the-Knot has a page devoted to this topic too.