# Why do the sines of the numerators of $\pi$’s continued fraction convergents approach zero?

I was messing with the sine function and tried getting values close to zero with integer inputs. I found a peculiar pattern. If you take pi’s continued fraction and write them out as one whole fraction, and take the numerators, you get the sequence here: http://oeis.org/A046947. If you take the sin (radians) of this sequence, you get values very very close to zero. For example, if you take the last listed number on oeis, the sin is only -2*10^-14. Is there any reason as to why this happens?

• By definition they get closer and closer to an integer (the corresponding denominator) multiple of $\pi$, closeness being of the order of the reciprocal of the denominator or better, so obviously the $\sin$ will be at most the reciprocal of this up to a constant like say 4, so for example the last term having 13 digits, you expect exactly something like $10^{-13}$ or so – Conrad Aug 17 '19 at 0:16
• Sometimes you can even do better like in the notable small excellent approximations like $\frac{22}{7}, \frac{355}{113}$ which are well known from antiquity where you get a smaller absolute value for the sin at the numerator than the expected reciprocal of the denominator (both above cases give you another $10^{-1}$) checking the values, while $333$ gives you about what you would expect – Conrad Aug 17 '19 at 0:31

Most denominators $$q$$ will get you a fraction $$p/q$$ that is within $$1/2q$$ of $$\pi$$. A convergent will get you a fraction within $$1/q^2$$. So $$p$$ is within $$1/q$$ of $$q\pi$$, and its sine is very small.