Showing $\frac{1-\cos \left((n-1)\pi/n\right)}{1-\cos \left(\pi/n\right)} \approx 4n^2/\pi^2$

Problem

Show $$\frac{1-\cos \left((n-1)\pi/n\right)}{1-\cos \left(\pi/n\right)} \approx 4n^2/\pi^2$$

Try

I have noticed that the numerator can be approximated

$$1-\cos \left((n-1)\pi/n\right) \approx 2$$

and the denominator can be approximated

\begin{aligned} 1 - \cos(\pi/n) &= (\pi/n)^2/2 - (\pi/n)^4/24 + \cdots \\ &\approx \pi^2 n^2/2 \end{aligned}

Thus we have the approximation

$$\frac{1-\cos \left((n-1)\pi/n\right)}{1-\cos \left(\pi/n\right)} \approx 4n^2/\pi^2$$

However, I feel dubious about this approximation assumes the independence of numerator and denominator, but they depend on each other through $$n$$.

So is there rigorous approach to here?

Any help will be appreciated.

To show that $$\dfrac{1-\cos ((n-1)\pi/n)}{1-\cos (\pi/n)} \approx 4n^2/\pi^2$$, note that $$\cos(2x) =\cos^2(x)-\sin^2(x) =1-2\sin^2(x)$$ so $$1-\cos(2x) =2\sin^2(x)$$ or $$1-\cos(x) =2\sin^2(x/2)$$.

Therefore

$$\begin{array}\\ \dfrac{1-\cos ((n-1)\pi/n)}{1-\cos (\pi/n)} &=\dfrac{2\sin^2((n-1)\pi/(2n))}{2\sin^2(\pi/(2n))} \\ &=\dfrac{\sin^2((1-1/n)\pi/(2))}{\sin^2(\pi/(2n))}\\ &\approx\dfrac{\sin^2(\pi/2-\pi/(2n))}{(\pi/(2n))^2} \qquad\text{since }\sin(x) \approx x \text{ for small } x\\ &=\dfrac{4n^2\cos^2(\pi/(2n))}{\pi^2}\\ &\approx\dfrac{4n^2}{\pi^2} \qquad\text{since } \cos(x) \approx 1 \text{ for small }x\\ \end{array}$$

We have $$\cos \left((n-1)\pi/n\right)=-\cos \left(\pi/n\right)$$ and then $$\frac{1-\cos \left((n-1)\pi/n\right)}{1-\cos \left(\pi/n\right)} =\frac{1+\cos \left(\pi/n\right)}{1-\cos \left(\pi/n\right)}= \frac{2\cos^2 \left(\pi/2n\right)}{2\sin^2 \left(\pi/2n\right)} = \left(\frac{1}{\tan \left(\pi/2n\right)}\right)^2 \approx \left(\frac{1}{\left(\pi/2n\right)}\right)^2=4n^2/\pi^2$$ where $$\tan x\approx x$$ for small $$x$$.

• Nice. Simpler than mine. – marty cohen May 7 at 4:13

Just for the fun

Starting from @Nosrati's answer, using Taylor expansions, binomial theorem and long division $$\tan \left(\frac{\pi }{2 n}\right)=\frac{\pi }{2 n}+\frac{\pi ^3}{24 n^3}+\frac{\pi ^5}{240 n^5}+O\left(\frac{1}{n^7}\right)$$ $$\tan ^2\left(\frac{\pi }{2 n}\right)=\frac{\pi ^2}{4 n^2}+\frac{\pi ^4}{24 n^4}+\frac{17 \pi ^6}{2880 n^6}+O\left(\frac{1}{n^8}\right)$$ $$\frac 1 {\tan ^2\left(\frac{\pi }{2 n}\right) }=\frac{4 n^2}{\pi ^2}-\frac{2}{3}+\frac{\pi ^2}{60 n^2}+O\left(\frac{1}{n^4}\right)$$

Using it for $$n=2$$, instead of $$1$$, we get $$-\frac{2}{3}+\frac{16}{\pi ^2}+\frac{\pi ^2}{240}\approx 0.995596$$ For $$=3$$ $$-\frac{2}{3}+\frac{36}{\pi ^2}+\frac{\pi ^2}{540}\approx 2.99917$$