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

Question

Problem

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

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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.

Answer 1

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} $

Answer 2

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 $x$.