Let $n$ be an integer, then:
$$\sin{nθ}=sinθ[\binom{n}{0}(2\cosθ)^{n-1}-\binom{n-1}{1}(2\cosθ)^{n-3}+\binom{n-2}{2}(2\cosθ)^{n-5}-...]$$
$$\cos{nθ}=\frac{1}{2}[(2cosθ)^{n}-\frac{n}{1}\binom{n-2}{0}(2\cosθ)^{n-2}+\frac{n}{2}\binom{n-3}{1}(2cosθ)^{n-4}-...]$$
You can get other identities by setting $$θ=\frac{π}{2}-ϕ; $$ and then consider different cases when n is even or odd and so on.
Either way, in the second series $\cos{nθ}$ is in terms of powers of cosines, set $\cos{nθ}$ an arbitrary value, say $p$, then we also have that $\cos{(nϕ+2π)},\cos{(nϕ+4π)},...$ satisfy the equation, hence $\cos{(ϕ)},\cos{(ϕ+\frac{2π}{n})},\cos{(ϕ+\frac{4π}{n})},...$ are the roots of the equation on the right hand side, there are exactly n roots. Let$$ \cosθ=\frac{1}{q}$$
Upong making this substitution multiply by $q^n$ on both sides of the identity, then the roots of the new equation become $\sec{(ϕ)},\sec{(ϕ+\frac{2π}{n})},\sec{(ϕ+\frac{4π}{n})},...$, I'll consider the case when $n$ is odd, then $$\cos{nθ}=2^{n-1}(cosθ)^{n}-\frac{n}{1}\binom{n-2}{0}2^{n-3}(\cosθ)^{n-2}+...+(-1)^{\frac{n-1}{2}}n\cosθ$$ Making the said substitution and multiplying by $q^n$ yields: $$q^n\cos{nθ}=2^{n-1}-\frac{n}{1}\binom{n-2}{0}2^{n-3}q^2+...+(-1)^{\frac{n-1}{2}}nq^{n-1}$$It is a well known fact that the sum of roots is equal to the coefficient of the $q^{n-1}$ term divided by the coefficient of $q^{n}$, therefore:
$$\sum_{k=1}^{n}\sec{(ϕ+\frac{(2k-2)π}{n})}=(-1)^{\frac{n-1}{2}}n\sec{nϕ}$$
Furthermore, $p_1^2+...+p_n^2=(p_1+...+p_n)^2-2\sum_{i≠j}p_ip_j$, but the sum of the roots taken two at a time is the coefficient of $q^{n-2}$ which is zero when $n$ is odd, thus when n is odd we have:
$$\sum_{k=1}^{n}\sec^2{(ϕ+\frac{(2k-2)π}{n})}=n^2\sec^2{nϕ}$$
A similar derivation goes when n is even, for the sum of cosecants you may want to expand the sine in powers of sines and make the natural substitution $\sinθ=\frac{1}{q}$ and then use similar arguments with the roots so that the sum of its roots is a known coefficient, furthermore you can let $\sin^2{θ}=\frac{1}{q}$ and then let $p=q-1$ because $\cot^2{θ}+1=\csc^2{θ}$ so that by the same argument you can get the sum of cotangets and also of the cotangets squared, in the same manner one uses the cosines to build up the secant and then make use of the secant/tangent identity to find the sum of tangents.
If you are interested in the the original series, you can derive them from the identity:
$$\frac{\sin{θ}}{1-2xcos{θ}+x^2}=\sinθ+x\sin{2θ}+x^2\sin{3θ}+...ad inf$$
and:
$$\frac{1-x^2}{1-2x\cos{θ}+x^2}=1+2x\cosθ+2x^2\cos{2θ}+... adinf$$
In both expressions you can expand the denominator as a geometric series if $x<1$ and the compare coefficients and finally get the expressions for $\sin{nθ}$ and $2\cos{nθ}$. (and by the way, the series in the beginning end whenever the binomial coefficient is either $\binom{n}{n-1}$ or $\binom{n}{n}$, and the signs alternate as plus, minus, plus,...)
As you can see, the mathematical analysis way is easier and more convenient whilst using trigonometry can be tedious, the coefficients of this series behave nicely using the binomial notation, I hope it doesn't bother you posting this 2 years later but I figured it out I could employ trigonometry to somehow solve your problem.
Best Answer
It would be naïve and incorrect to proceed as follows
$$\begin{align} \sum_{k=1}^{n-1}\frac{\cot^2(\pi k/n)}{n^2}&\underbrace{\approx}_{\text{WRONG!}} \frac1n\int_{1/n}^{1-1/n}\cot^2(\pi x)\,dx\\\\ &=\frac1n\left.\left(-x-\frac1\pi \cot(\pi x)\right)\right|_{1/n}^{1-1/n}\\\\ &=\frac2{n^2}-\frac1n +\frac2{n\pi}\cot(\pi/n)\\\\ &\to \frac2{\pi^2} \end{align}$$
Instead, we use $\cot^2(x)=\csc^2(x)-1$ to write
$$\begin{align} \sum_{k=1}^{n-1}\frac{\cot^2(\pi k/n)}{n^2}&=\frac1n-\frac1{n^2}+\sum_{k=1}^{n-1}\frac{1}{n^2\,\sin^2(\pi k/n)}\\\\ &=\frac1n-\frac1{n^2}+2\sum_{k=1}^{\lfloor n/2\rfloor-1}\frac{1}{n^2\,\sin^2(\pi k/n)} \end{align}$$
Next, we note that for $\pi/2>x>0$, $\left(x-\frac16 x^3\right)^2\le \sin^2(x)\le x^2$. Hence, we have
$$\begin{align} \frac2{\pi^2}\sum_{k=1}^{\lfloor n/2\rfloor-1}\frac1{k^2} \le \frac2{n^2}\sum_{k=1}^{\lfloor n/2\rfloor-1}\frac{1}{n^2\,\sin^2(\pi k/n)}&\le \frac2{\pi^2}\sum_{k=1}^{\lfloor n/2\rfloor-1}\frac1{k^2\left(1-\frac16\frac{\pi^2k^2}{n^2}\right)^2}\\&=\frac2{\pi^2}\sum_{k=1}^{\lfloor n/2\rfloor-1}\frac1{k^2}+O\left(\frac1n\right) \end{align}$$
whence letting $n\to \infty$ and applying the squeeze theorem yields the coveted limit
$$\lim_{n\to\infty}\sum_{k=1}^{n-1}\frac{\cot^2(k\pi/n)}{n^2}=\frac13$$