$n-$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.
 A: Well, with very large values of $n$, you will get
$$\sin(x)=\text{polynomial in $\sin\left(\frac{x}{n}\right)$ or $\cos\left(\frac{x}{n}\right)$  or both with a very large degree}$$
where the degree of a polynomial is the highest power in it, for example the degree of $x^2+3x+5$ is $2$.
However, we don't general formulae for any general polynomial equation with degree greater than or equal to $5$ then I guess for $n\ge5$ what you ask is indeed impossible.
I have no rigorous response to Bumblebee's point, but the following formulae makes it seem unlikely for your question to have 'yes, it's possible at least sometimes (apart from the trivial cases)' for an answer.
$$\begin{align}
\sin(n\theta) &= \sum_{k\text{ odd}} (-1)^\frac{k-1}{2} {n \choose k}\cos^{n-k} \theta \sin^k \theta, \\
\cos(n\theta) &= \sum_{k\text{ even}} (-1)^\frac{k}{2} {n \choose k}\cos^{n-k} \theta \sin^k \theta \,,
\end{align}$$
A: 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
$$\dfrac{\cos\left(\dfrac{\theta}{n}\right)}{\sqrt[n]{\cos\theta}}=1+\dfrac{(n-1)}{2!n^2}\tan^2\theta-\dfrac{(n-1)(2n-1)(3n-1)}{4!n^4}\tan^4\theta +\cdots.$$
A: This is not a complete answer but I think it will be too long for a comment and moreover it gives you a explicit formula for the case when $n$ is a power of $2$:
For a given $\alpha$, let's call $\beta=\alpha/2$. Then your formula applied to $\beta/2$ reads
$$
\sin\left(\frac{\beta}{2}\right) = \pm\sqrt{\frac{1-\cos(\beta)}{2}}.
$$
But $\beta=\alpha/2$ so we can apply agian the same formula:
$$
\sin\left(\frac{\alpha}{4}\right) = \pm\sqrt{\frac{1-\cos\left(\frac{\alpha}{2}\right)}{2}}= \pm\sqrt{\frac{1\mp\sqrt{\frac{1-\cos(\alpha)}{2}}}{2}} =\pm\sqrt{\frac{1\mp\sqrt{\frac{1-\cos(\alpha)}{2}}}{2}} .
$$
It can be simplified a little as
$$
\sin\left(\frac{\alpha}{4}\right) = \pm\sqrt{\frac{\sqrt 2 \mp\sqrt{1-\cos(\alpha)}}{2\sqrt 2}} .
$$
You can play the same game for $\gamma=\beta/2$. Undoing the change of variable, and if I am not mistaken, the formula should be
$$
\sin\left(\frac{\alpha}{8}\right) = \pm\sqrt{\frac{2\sqrt2 \mp\sqrt{\sqrt2\mp \sqrt{1-\cos(\alpha)}}}{2\sqrt{2\sqrt2}}} .
$$
To get the general formula, it is better to denote square roots by $2^{1/2}$, so that $\sin(\alpha/8)$ becomes
$$
\sin\left(\frac{\alpha}{8}\right) = \pm\sqrt{\frac{2^{1+1/2} \mp\sqrt{2^{1/2}\mp \sqrt{2^0-\cos(\alpha)}}}{2^{1+1/2+1/4}}} .
$$
And there you go:
$$
\sin\left(\frac{\alpha}{2^m}\right) = \pm\sqrt{\frac{2^{2^{1-(m-1)}\left(2^{m-1})-1\right)} \mp\sqrt{2^{2^{1-(m-2)}\left(2^{m-2})-1\right)} \mp \sqrt{\cdots \sqrt{1-\cos(\alpha)}}}}{2^{2^{1-m}\left(2^m-1\right)}}} .
$$
Remark. I am using the fact that
$$
\sum_{k=0}^{m-1} \frac{1}{2^k} = 2^{1-m}\left(2^m-1\right)
$$
