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.

triangle

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.

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Responses

  1. 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.

    • Nice! Thanks.

  2. 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

  3. cos theta = – 1/ sqrt(2) and not 1/sqrt(2)

  4. Cosine value should be negative to make the resulting angle 3pi/4.

    • Right—thanks! Good catch.

  5. If tanA, tanB, and tanC are nonzero, then tanA + tanB + tanC = tanA*tanB*tanC if and only if A + B + C = Pi*K, where k is some integer. This together with the fact that 1 + 2 + 3 = 1*2*3 immediately gives arctan1 + arctan2 + arctan3 = Pi.


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