Basis formed by hyperbolic functions I am currently working with separation of variables for different kinds of PDEs and one often uses here the fact that one has the sine base, i.e.,
$$ \left( \sin(k\pi y) \right)_{k=1}^{\infty} $$
forms a base of $L^2(0,1)$. This also holds on discrete level, i.e., having vector $v_k \in \mathbb{R}^{n-1}$ defined entry-wise as
$$ v_k = \left( \sin(\frac{k\pi}{n} j) \right)_{j=1}^{n-1} $$
then the set $\{v_k\}_{k=1}^{n-1}$ is basis of $\mathbb{R}^{n-1}$.
However, in some cases it would be much more suitable to use the hyperbolic functions - in this case the hyperbolic sine. My question is whether or not the analogue holds for hyperbolic sines as well, i.e., 
(A) Is the $\left( \sinh(k\pi y) \right)_{k=1}^{\infty}$ basis of some reasonable Lebesgue space on $(0,1)$?
(B) Having vector $w_k \in \mathbb{R}^{n-1}$ defined entry-wise as
$$ w_k = \left( \sinh(\frac{k\pi}{n} j) \right)_{j=1}^{n-1}, $$
is the set $\{w_k\}_{k=1}^{n-1}$ a basis of $\mathbb{R}^{n-1}$?
Update: The computations suggest that also the vectors $w_k$ form a basis, but one that is incresingly ill-conditioned when $n$ grows. For $n=100$, python computed that the condition number is of order $10^{130}$. I am aware that at that point it is impossible to argue with such a result, so the question remains open. Also, I am interested only in the theoretical usage and I do not intend to use this basis (if it indeed is one in general) for computational purposes.
 A: For (B), the answer is affirmative. Actually the determinant is easy to compute. Let
$$D(z_1,\ldots,z_n)=\det\{z_i^j-z_i^{-j} : 1\leqslant i,j\leqslant n\}$$
with nonzero $z_1,\ldots,z_n\in\Bbb{C}$. Fix $z_1,\ldots,z_{n-1}$ so that $1,z_1,\ldots,z_{n-1},-1,1/z_1,\ldots,1/z_{n-1}$ are all pairwise distinct. We have
$$D(z_1,\ldots,z_{n-1},z)=\sum_{k=1}^{n}a_k(z^k-z^{-k})=z^{-n}P(z)$$
(expanding along the last row), where $a_k$ (are $\pm$ lower-sized determinants that) do not depend on $z$, and thus $P(z)$ is a polynomial of degree $2n$ with leading coefficient $A=D(z_1,\ldots,z_{n-1})$. Further, $P(1)=P(-1)=0$ (row of zeros) and $P(z_k)=P(1/z_k)=0$ for $0<k<n$ (equal rows). This shows that $P(z)=A(z^2-1)\prod_{k=1}^{n-1}(z-z_k)(z-1/z_k)$, and we obtain
$$D(z_1,\ldots,z_n)=\prod_{k=1}^{n}\left[\Big(z_k-\frac{1}{z_k}\Big)\prod_{r=1}^{k-1}\Big(z_k+\frac{1}{z_k}-z_r-\frac{1}{z_r}\Big)\right].$$
As a corollary, we get
$$\det\{\sinh ijz : 1\leqslant i,j\leqslant n\}=2^{n(n-1)/2}\prod_{k=1}^{n}\Big(\sinh kz\prod_{r=1}^{k-1}(\cosh kz-\cosh rz)\Big).$$
For (A), the linear span of this system is dense in $L_2(0,1)$ (for a proof, it can be shown first that it is dense in $\{f\in C([0,1]) : f(0)=0\}$; this can be done using some form of the Stone-Weierstrass theorem). Thus, this system becomes a Hilbert basis of $L_2(0,1)$ after orthonormalization (the system itself does not have this property — this can be shown similarly to the case of the system of monomials — but, of course, yes, it is a basis of some space ;).
