# 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.

1. Michael Lugo says:

I think of this in terms of complex numbers: (1+2i)(1+3i) = -5 + 5i. Taking arguments of the complex numbers gives arg (1+2i) + arg(1+3i) = arg(-5 + 5i). Recalling that arg(a+bi) = arctan (b/a) gives what you’re looking for.

1. Nice! Thanks.

2. lzyerste says:

tan(A+B) = (tan(A) + tan(B)) / (1 – tan(A) * tan(B))
So, tan(arctan(2) + arctan(3)) = (2 + 3) / (1 – 2 * 3) = -1
=> arctan(2) + arctan(3) = (3/4) * pi