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.
 A: 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 $x$.
A: 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}
$
