In a previous post I defined two functions that appear in the discrete Hartley transforms. Here, I will re-define them with a different normalization:
\begin{align*}
\operatorname{cps}{(x)} &\equiv \frac{\cos{(x)} + \sin{(x)}}{\sqrt{2}}, & \operatorname{cqs}{(x)} &\equiv \frac{\cos{(x)} - \sin{(x)}}{\sqrt{2}}
\end{align*}
Let us look at a few identities.
You have the parity/reflection identities:
\begin{align*}
\operatorname{cps}{(-x)} &= \operatorname{cqs}{(x)}, & \operatorname{cqs}{(-x)} &= \operatorname{cps}{(x)}
\end{align*}
You have argument addition:
\begin{align*}
\sqrt{2}\operatorname{cps}{(x + y)} &= \operatorname{cps}{(x)} \operatorname{cps}{(y)} + \operatorname{cps}{(x)} \operatorname{cqs}{(y)} + \operatorname{cqs}{(x)} \operatorname{cps}{(y)} - \operatorname{cqs}{(x)} \operatorname{cqs}{(y)} \\
\sqrt{2}\operatorname{cqs}{(x + y)} &= \operatorname{cqs}{(x)} \operatorname{cqs}{(y)} + \operatorname{cqs}{(x)} \operatorname{cps}{(y)} + \operatorname{cps}{(x)} \operatorname{cqs}{(y)} - \operatorname{cps}{(x)} \operatorname{cps}{(y)}
\end{align*}
You have argument subtraction:
\begin{align*}
\sqrt{2}\operatorname{cps}{(x - y)} &= \operatorname{cps}{(x)} \operatorname{cps}{(y)} + \operatorname{cps}{(x)} \operatorname{cqs}{(y)} - \operatorname{cqs}{(x)} \operatorname{cps}{(y)} + \operatorname{cqs}{(x)} \operatorname{cqs}{(y)} \\
\sqrt{2}\operatorname{cqs}{(x - y)} &= \operatorname{cqs}{(x)} \operatorname{cqs}{(y)} + \operatorname{cqs}{(x)} \operatorname{cps}{(y)} - \operatorname{cps}{(x)} \operatorname{cqs}{(y)} + \operatorname{cps}{(x)} \operatorname{cps}{(y)}
\end{align*}
The product of the two functions has a simple form:
\begin{equation*}
2\operatorname{cps}{(x)}\operatorname{cqs}{(x)} = \cos{(2x)}
\end{equation*}
Taking squares gives
\begin{align*}
\operatorname{cps}^{2}{(x)} &= \frac{1 + \sin{(2x)}}{2} & \operatorname{cqs}^{2}{(x)} &= \frac{1 - \sin{(2x)}}{2}
\end{align*}
Thus,
\begin{align*}
\operatorname{cps}^{2}{(x)} + \operatorname{cqs}^{2}{(x)} &= 1, & \operatorname{cps}^{2}{(x)} - \operatorname{cqs}^{2}{(x)} &= \sin{(2x)}
\end{align*}
The ratio of the two functions also has a simple form, in terms of the tangent function:
\begin{equation*}
\frac{\operatorname{cps}{(x)}}{\operatorname{cqs}{(x)}} = \frac{1 + \tan{(x)}}{1 - \tan{(x)}}
\end{equation*}