the orbit functions into polynomials is $y=e^{2\pi i x}$, monomials in such a polynomial are $y^m$ and $y^{-m}$. Instead, we introduce new (‘trigonometric’) variables $X$ and $Y$ as follows: $$\begin{gathered}
X:=C_1(x){=}e^{2\pi i x}{+}e^{{-}2\pi i x}{=}2\cos(2\pi x),\label{def_X}\\
Y:=S_1(x){=}e^{2\pi i x}{-}e^{{-}2\pi i x}{=}2i\sin(2\pi x).\label{def_Y}\end{gathered}$$ We can now start to construct polynomials recursively in the degrees of $X$ and $Y$, by calculating the products of the appropriate orbit functions. Omitting the dependence on $x$ from the symbols, we have $$\begin{gathered}
\begin{array}{rlcrl}
X^2 \!\!\!&=C_2+2             &\Longrightarrow\quad&     C_2\!\!\!&=X^2-2,\\
XC_2\!\!\!&=C_3+X             &\Longrightarrow\quad&     C_3\!\!\!&=X^3-3X,\\
XC_m\!\!\!&=C_{m+1}+C_{m-1}   &\Longrightarrow\quad& C_{m+1}\!\!\!&=XC_m-C_{m-1},\quad m\geq3.
\end{array}\label{C_via_X}\end{gathered}$$ Therefore, we obtain the following recursive polynomial form of the $C$-functions $$\begin{gathered}
\label{C-form_of_Tn}
C_0=2,\quad
C_1=X,\quad
C_2=X^2{-}2,\quad
C_3=X^3{-}3X,\quad
C_4=X^4{-}4X^2+2,\dots .\end{gathered}$$ After the substitution $z=\tfrac 12 X$ we have $$\begin{gathered}
C_0{=}2\cdot1,\quad
C_1{=}2z,\quad
C_2{=}2(2z^2{-}1),\quad
C_3{=}2(4z^3{-}3z),\quad
C_4{=}2(8z^4{-}8x^2+1),\dots.\end{gathered}$$ Hence we conclude that $C_m\!=2T_m$, for $m=0,1,\dots$.

In our opinion, the normalization of orbit functions is also more ‘natural’ for the Chebyshev polynomials. For example, the equality $C_2^2=C_4+2$ does not hold for $T_2$ and $T_4$.

Each $C_m$ also can be written as a polynomial of degree $m$ in $X$,$Y$ and $S_{m-1}$. It suffices to consider the products $YS_m$, e.g., $C_2=Y^2+2$, $C_3=YS_2+X$, etc. Equating the polynomials obtained in such a way with the corresponding polynomials from (\[C\_via\_X\]), we 