$n-$section formulas in goniometry without the calculus We know that the formulas of bisection of an angle $\alpha$ it is:
$$\sin^2\left(\frac{\alpha}2\right)=\frac{1-\cos(\alpha)}{2}, \qquad \cos^2\left(\frac{\alpha}2\right)=\frac{1+\cos(\alpha)}{2}$$
In particular
$$\sin\left(\frac{\alpha}2\right)=\pm\sqrt{\frac{1-\cos(\alpha)}{2}}, \qquad \cos\left(\frac{\alpha}2\right)=\pm\sqrt{\frac{1+\cos(\alpha)}{2}}$$
My question is:
For the formulae of $n-$section of an angle $\alpha$, i.e.
$$\sin\left(\frac{\alpha}n\right), \qquad \cos\left(\frac{\alpha}n\right)$$
Do I must necessarily use the development in series of Taylor or the general formulas can be obtained with methods that do not use the derivatives, or the numerical methods?
My doubts arise for the students of an high school that not use the derivate.

I doubt that there are such elementary general formulas. I will derive a possible (restricted) formula as an infinite series using De Moivre's formula, though this may not satisfy your requirement of calculus free derivation. One value of the multivalued expression $\cos\left(\dfrac{\theta}{n}\right)+i\sin\left(\dfrac{\theta}{n}\right),\quad n\ge2$ is given by
\begin{align}
\left(e^{i\theta}\right)^{1/n} & 
= \left(\cos\theta+i\sin\theta\right)^{1/n} \\
 & = \left(\cos\theta\right)^{1/n}\left(1+i\tan\theta\right)^{1/n}
\end{align}
Now, assuming $0\lt\theta\lt\pi/4,$ we can expand $\left(1+i\tan\theta\right)^{1/n}$ via binomial series as $$1+\dfrac{1}{n}i\tan\theta +\dfrac{(n-1)}{2!n^2}\tan^2\theta -\dfrac{(n-1)(2n-1)}{3!n^3}i\tan^3\theta -\dfrac{(n-1)(2n-1)(3n-1)}{4!n^4}\tan^4\theta +\cdots.$$
Hence we have two infinite series $$\dfrac{\sin\left(\dfrac{\theta}{n}\right)}{\sqrt[n]{\cos\theta}}=\dfrac{1}{n}\tan\theta - \dfrac{(n-1)(2n-1)}{3!n^3}\tan^3\theta+ \dfrac{(n-1)(2n-1)(3n-1)(4n-1)}{5!n^5}\tan^5\theta+$$ and
$$\dfrac{\cos\left(\dfrac{\theta}{n}\right)}{\sqrt[n]{\cos\theta}}=1+\dfrac{(n-1)}{2!n^2}\tan^2\theta-\dfrac{(n-1)(2n-1)(3n-1)}{4!n^4}\tan^4\theta +\cdots.$$
