obtain a trigonometric identity for each $m$. For example, we find two ways to write $C_2$, one from the product $X^2$ and one from $Y^2$. Equating the two, we get $$\begin{gathered}
X^2{-}Y^2=4\quad\Longleftrightarrow\quad \sin^2(2\pi x){+}\cos^2(2\pi x)=1\end{gathered}$$ because $Y$ is defined in (\[def\_Y\]) to be purely imaginary.

Just as the polynomials representing $C_m$ were obtained above, it is possible to to find polynomial expressions for $S_m$ for all $m$.

Fundamental relations between the $S$- and $C$- orbit functions follow from the properties of the character $\chi_m(x)$ of the irreducible representation of $A_1$ of dimension $m+1$.

The character can be written in two ways: as in the Weyl character formula and also as the sum of appropriate $C$-functions. Explicitly, we have the $A_1$ character: $$\begin{gathered}
\chi_m(x)=\frac{S_{m+1}(x)}{S_1(x)}=C_m(x)+C_{m-2}(x)+\cdots+
    \begin{cases}
    C_2(x)+1\quad    &\text {for $m$ even},\\
    C_3(x)+C_1(x)\quad &\text {for $m$ odd}.
    \end{cases}\end{gathered}$$ Let us write down a few characters $$\begin{gathered}
\chi_0=\tfrac{S_{1}(x)}{S_1(x)}=1,\quad
\chi_1=\tfrac{S_{2}(x)}{S_1(x)}=C_1=X,\quad
\chi_2=\tfrac{S_{3}(x)}{S_1(x)}=C_2+C_0=X^2-1,\\
\chi_3=\tfrac{S_{4}(x)}{S_1(x)}=C_3+C_1=X^3-2X,\quad
\chi_4=\tfrac{S_{5}(x)}{S_1(x)}=C_4+C_2+C_0=X^4-3X^2+1,\dots\end{gathered}$$ Again, the substitution $z=\tfrac 12 X$ transforms these polynomials into the Chebyshev polynomials of the second kind $\frac{S_{m+1}}{S_1}=U_m$,  $m=0,1,\dots$, indeed $$\begin{gathered}
\tfrac{S_{1}(x)}{S_1(x)}=1,\quad
\tfrac{S_{2}(x)}{S_1(x)}=2z,\quad
\tfrac{S_{3}(x)}{S_1(x)}=4z^2{-}1,\quad
\tfrac{S_{4}(x)}{S_1(x)}=8z^3{-}4z,\quad
\tfrac{S_{5}(x)}{S_1(x)}=16z^4{-}12z^2{+}1,\dots\end{gathered}$$

Note that in the character formula we used $C_0=1$, while above (see (\[def\_e-function2\]) and (\[C-form\_of\_Tn\])) we used $C_0=2$. It is just a question of normalization of 