I try to understand better the parametrisation of the $n-$sphere (suppose $R=1$). I know that it is given in $\mathbb R^{n+1}$ by $$\begin{cases}\cos\theta _n\cos\theta _{n-1}...\cos\theta _1\\ \cos\theta _n...\cos\theta _2\sin\theta _1\\ \cos\theta _n...\cos\theta _3\sin\theta _2\\ \vdots \\\cos \theta _n\sin\theta _{n-1}\\ \sin\theta _n\end{cases},\quad \theta _i\in [0,2\pi[, i=1,...,n-1, \theta _n\in [-\pi/2,\pi/2]$$
So, for $n=1$, it is just $(\cos\theta ,\sin\theta )$ with $\theta \in [0,2\pi]$ and in cartesian is $x^2+y^2=1$. For $n=2$, we have have in cartesian $$x^2+y^2+z^2=1,$$ So by setting $u^2=x^2+y^2$, we have half circle in $Ouz$ parametrized by $(u,z)=(\cos\theta ,\sin\theta )$ with $\theta \in [-\pi/2,\pi/2]$ and a circle in $Oxy$ parmetrized by $(u\cos\varphi,u\sin\varphi)$ with $\varphi\in [0,2\pi[$. At the end, in $Oxyz$, we get $$(x,y,z)=(\cos\theta \cos\varphi,\cos\theta \sin \varphi,\sin \theta )$$ with $\varphi\in [0,2\pi]$ and $\theta \in [-\pi/2,\pi/2]$.
I know that the cartesian equation of an hypersphere is given by $$x_1^2+\cdots+x_n^2=1.$$ Let take $n=4$, so $$x_1^2+\cdots+x_4^2=1.$$ If I follow the previous construction, we should have a sphere on $Ox_1x_2x_3$ and a half circle on $Oux_4$, where $u^2=x_1^2+x_2^2+x_3^2$. Therefore, $(u,x_4)=(\cos \theta _1,\sin\theta _1)$ with $\theta _1\in [-\pi,\pi]$ and $$(x_1,x_2,x_3)=(u\cos\theta \cos\varphi,u\cos\theta \sin\varphi,u\sin\theta ),\quad \theta \in [-\pi/2,\pi/2],\varphi\in [0,2\pi],$$ and thus $$(x_1,x_2,x_3,x_4)=(\cos\theta _1\cos\theta \cos\varphi,\cos\theta _1\cos\theta \sin\varphi,\cos\theta _1\sin\theta,\sin\theta _1 ),\quad \theta _1,\theta \in [-\pi/2,\pi/2],\varphi\in [0,2\pi].$$
We can then continue like this, and thus, we should obtain
$$\begin{cases}\cos\theta _n\cos\theta _{n-1}\cdots\cos\theta _1\\ \cos\theta _n\cdots\cos\theta _2\sin\theta _1\\ \cos\theta _n\cdots\cos\theta _3\sin\theta _2\\ \vdots \\\cos \theta _n\sin\theta _{n-1}\\ \sin\theta _n\end{cases},\quad \theta _1\in [0,2\pi[, \theta _i\in [-\pi/2,\pi/2], i=2,\cdots,n.\tag{E}$$
Question 1 : Why such a construction does not work ?
Question 2 : What would be the surface obtained by $(E)$?
Question 3 : Finally, how is constructed the $n$-sphere?
In fact, for $n \geqslant 3$, the range of parameters given at the beginning is incorrect whereas that in (E) is correct. To see this, consider an arbitrary point $(x_1, \cdots, x_{n + 1}) \in S^n$ such that $x_1, \cdots, x_{n + 1} ≠ 0$. Suppose$$ \begin{cases} x_1 = \cos θ_1 \cos θ_2 \cdots \cos θ_n\\ x_2 = \sin θ_1 \cos θ_2 \cdots \cos θ_n\\ x_3 = \sin θ_2 \cos θ_3 \cdots \cos θ_n\\ \vdots\\ x_{n + 1} = \sin θ_n \end{cases}, $$ where $θ_1, \cdots, θ_{n - 1} \in [0, 2π)$, $θ_n \in \left[\dfrac{π}{2}, \dfrac{π}{2}\right]$. Now take$$ (θ_1', θ_2', θ_3', \cdots, θ_n') = \begin{cases} (θ_1 + π, π - θ_2, θ_3, \cdots, θ_n), & 0 < θ_1 < π,\ 0 < θ_2 < π\\ (θ_1 - π, π - θ_2, θ_3, \cdots, θ_n), & π < θ_1 < 2π,\ 0 < θ_2 < π\\ (θ_1 + π, 3π - θ_2, θ_3, \cdots, θ_n), & 0 < θ_1 < π,\ π < θ_2 < 2π\\ (θ_1 - π, 3π - θ_2, θ_3, \cdots, θ_n), & π < θ_1 < 2π,\ π < θ_2 < 2π \end{cases}, $$ then $(θ_1', \cdots, θ_n') \mapsto (x_1, \cdots, x_{n + 1})$.