A couple days ago I saw this tweet.

Pretty cool! Let’s see why

Two terms are easy to deal with:

and

But why is

One way to prove this is via the inverse trigonometric identity

In this case

Because and are between 0 and , .

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 in the diagram below.

The law of cosines tells us that

This implies that and hence

(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.

### Like this:

Like Loading...

*Related*

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.

By:

Michael Lugoon September 9, 2013at 3:35 pm

Nice! Thanks.

By:

Dave Richesonon September 9, 2013at 3:37 pm

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

By:

lzyersteon September 30, 2013at 1:58 pm

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

By:

kali prasad tripathyon November 4, 2013at 8:49 am

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

By:

John Molokachon January 1, 2014at 3:20 pm

Right—thanks! Good catch.

By:

Dave Richesonon January 26, 2014at 2:43 pm

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.

By:

Shahen Mirzoyanon September 19, 2014at 5:49 pm