Thursday, January 23, 2014

Math Challenge #1

Here's the first of some math challenges, drawn from my notebooks. This is one from around 1977.

Observe that sin(333) + sin(355) = sin(22) is not an equality, but is true to about 9 significant digits. Explain. Find another similar almost-identity.


Arlo said...

Well, if you think about it as a Taylor series it isn't so hard. 22/2*pi is very close to 3.5, likewise 333/2pi is 52.999 and 355/2pi is 56.000...

So each of these have second order Taylor approximations of: 0 +/- ~1*(n - (pi*N))

Where n is a small displacement and N is an integer.

Randy said...

Interesting. First thing I note is that the angles must be in radians. It doesn't work in degrees, even though the numbers appear to be in the interval suitable for degrees.

So, since π is important in radians, what do these look like in π?

355 ≈ 113.00π
333 ≈ 106.00π
22 ≈ 7.00π

sin θ ≈ 0 where θ ≈ kπ, k∈ℤ

sin repeats itself every 2π
|sin| repeats itself every π

Further, because sin is nearly vertical at these locations, we can note:

sin ε ≈ ε where ε is "small"

Let's keep 10 digits after the decimal (so we can round to 9) and separate out the whole π pieces.

(I just used subtraction of the above approximated values to get the decimal digits)

355 ≈ 113π + 0.000 030 144 4
333 ≈ 106π - 0.008 821 280 5
22 ≈ 7π + 0.008 851 424 9

sin 355 ≈ sin (113π + 0.000 030 144 4)
sin 355 ≈ -sin 0.000 030 144 4
sin 355 ≈ -0.000 030 144 4

sin 333 ≈ sin (106π - 0.008 821 280 5)
sin 333 ≈ -sin 0.008 821 280 5
sin 333 ≈ -0.008 821 280 5

sin 333 + sin 355 ≈ -0.008 851 424 9

sin 22 ≈ sin (7π + 0.008 851 424 9)
sin 22 ≈ -sin 0.008 851 424 9
sin 22 ≈ -0.008 851 424 9

What is "similar"? It seems we're required to find unique whole numbers x,y,z such that sin x + sin y ≈ sin z. My guess is there aren't enough in the set {0,1,2,...,359} to appear to be degrees.

If we're looking for small error values, approximations of π seem to be a good place to start.

22/7 has been used.
223/71 is available, but has an error large enough that I don't want to deal with it.
333/106 has been used.
355/133 has been used.
I'll use these approximations (from Wolfram MathWorld and

There's probably a better way, but trial and error produces this almost-identity, good to 15 or so digits:

sin 103993 + sin 208341 ≈ sin 104348
(I can't bring myself to use =)

Pseudonym said...

sin 355 = cos 22 sin 333 + cos 333 sin 22

Since and 333 is approximately 106π and 22 is approximately 7π (these are from the continued fraction expansion of π), the result more or less follows.

The next one in the series is (and again, it's not a true equality):

sin 103993 + sin 104348 = sin 355

Randy said...

There's always something more after I press Publish.

I forgot to mention there's a second requirement:

sin x + sin y ≈ sin z
x + z = y (note the slightly different order)

I should have written:

sin 103993 + sin 208341 ≈ sin 104348
103993 + 104348 = 208341

Nevertheless I lack a good way to produce these numbers other than looking them up. It may be that there are other good answers using the values in {0,1,2,...,359}.

Jeffrey Shallit said...

All answers so far are good, but incomplete. Pseudonym got the farthest, but even his/her answer leaves a bit to be desired.

Commenters noticed that the numbers 333, 355, and 22 are the numerators of good rational approximations to Pi. But how do we find such good approximations? They come from the simple continued fraction [3,7,15,1,292, 1, ...] for Pi. The terms of the continued fraction are called partial quotients.

The numerators are, consecutively, 3, 22, 333, 355, 103993, 104348, ... But while sin(333)+sin(355) is very close to sin(22), it's not the case that that sin(355)+sin(103993) is close to sin(333). So it's not simply that these are three consecutive convergents to Pi.

The solution is to take the numerators of three consecutive convergents to Pi,
say p, q, r where the partial quotient associated with r is 1 and occurs in an odd position. Then we have p + q = r and we also have p/p', q/q', and r/r' are all close to Pi.
Now use the addition law for sines: sin(p) = sin(r-q) = sin(r) cos(q) - sin(q) cos(r). Since r/r' is an odd convergent, r/r' is slightly greater than Pi, so cos(r) is very close to -1. Since q/q' is an even convergent, q/q' is slightly less than Pi, so cos(q) is very close to 1. Thus we get that sin(p) is very close to sin(r) + sin(q). So, for example, another pseudo-identity is sin(103993)+sin(104348) = sin(355).

If the 1 appears in an even position in the continued fraction, then we get a similar pseudo-identity with the order scrambled slightly: e.g.,
sin(104348) is roughly sin(103993)+sin(208341).

Jeffrey Shallit said...

Of course, there is a stupidity in what I wrote. We also need to take into account the parity of the denominators. In particular I should have said "Since r/r' is a convergent and r' is odd, r/r' is close to an odd multiple of Pi, so cos(r) is very close to -1; since q' is even, cos(q) is very close to 1."