Approximation of trig and polynomial function

Question

Find the smallest value of n for which $\frac{n}{\pi}\sin(\frac{\pi}{n}) \gt 0.999$

Not really sure where to start with this. The answer is 41. The question is from the Cambridge Y12 Ext-1 3U HSC textbook, exercise 9A Q20 b)

Answer 1

You didn't tell us how much calculus you've had. (Sorry, those Cambridge codes are over my head.) Where I grew up, in a small country in Eastern Europe, a long time ago, during communism, this would have been learnt in highschool. Later, in the U.S., I saw that the same would be taught in graduate courses at University.

So: $\sin x \approx x - \frac{x^3}6$ (Taylor series expansion). From this, heuristically, you get $\frac{\sin x}x \approx 1 - \frac{x^2}6$, and then you can prove rigorously that $\frac{\sin x}{x} > 1 - \frac{x^2}6$ for small, positive values of $x$. Now you want $n$ so that $x := \frac{\pi}n$ satisfies $\frac{x^2}6 < 0.001$, and solving for this gives $n > 40.55$ or so. Now, one would also need to show that $n = 40$ won't do, because we only used an inequality in the first place, not an equality. If you are allowed to use an electronic calculator for that, that's fine, but otherwise I don't know how one would prove that (only with pencil and paper).

Answer 2

Do we know that: $\sin x \approx x - \frac {x^3}{6} = x(1-\frac {x^2}{6})$

$\sin \frac {\pi}{n}= (1-0.001)\frac {\pi}n$

$(\frac {\pi}{n})^2 \approx 0.006\\ n \approx \frac {\pi}{\sqrt {0.006}}\\ n \approx 40.6$

$41$ would be the smallest $n\in N$

$x(1-\frac{x^2}{6})<\sin x<x(1-\frac{x^2}{6} +\frac{x^4}{120})$

$1-\frac{x^2}{6} + \frac {x^4}{120}= 1-0.001\\ x^4 - 20 x^2 + 0.12 = 0\\ x^2 = 10 - \sqrt {100-12}$

$0.0060018<(\frac {\pi}{n})^2 < 0.006\\ \frac {\pi}{\sqrt {0.0060018}}<(\frac {\pi}{n})^2 < \frac {\pi}{\sqrt {0.006}}$

$40.5517<n<40.5577$