# Rational Approximations Of Pi?

Computer Science Level pending

We have $$\sin(\pi/2 + 2\pi k) = 1$$ for all integers $$k.$$

Because $$\pi$$ is irrational, one can use this to show that $$\sin(m)$$ gets arbitrarily close to $$1$$ as $$m$$ ranges over all integers. Find the integer $$m$$ such that $$|m|\leq 1000$$ and $$\sin(m)$$ is as close as possible to $$1.$$

Details and assumptions

In other words, the answer to this question is the unique integer $$m$$ such that $$-1000\leq m\leq 1000$$ and $$1-\sin(m) \leq 1-\sin(n)$$ for all integers $$-1000\leq n\leq 1000$$.

×