What is $\sin(nx)$ iteration in terms of $\sin A$ and $\cos A$? I want to use sum angle formulas, $\sin(A+B)=\sin A\cos B+\cos A\sin B$
to get for any angles, $\sin(nA)$ in terms of powers of $\sin A$ and $\cos A$.

I know there are other ways, but I want to use trigonometry and iteration on that. The goal is to carry out iteration in terms of $\sin(nA)$?

 A: Is this OK? Trigonometry Sine Multiple Angle Formulae from Ken Ward's Mathematics Pages.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
You can get a recursive relation for $\ds{\sin\pars{nx}}$ because:
$$
\sin\pars{\bracks{m + 1}x} + \sin\pars{\bracks{m - 1}x} = 2\sin\pars{mx}\cos\pars{x}
$$
Set $\ds{m = n - 1}$ in the above expression to find:
\begin{align}
\bbox[#ffd,10px,border:1px dotted navy]{\sin\pars{nx}} & = -\sin\pars{\bracks{n - 2}x} +
2\sin\pars{\bracks{n - 1}x}\cos\pars{x}
\\[5mm] & =\
\bbox[#ffd,10px,border:1px dotted navy]{%
\sum_{k = 1}^{2}a_{k}\pars{x}\sin\pars{\bracks{n - k}x}}
\qquad\mbox{where}\qquad 
\left\{\begin{array}{rcl}
\ds{a_{1}\pars{x}} & \ds{=} & \ds{2\cos\pars{x}}
\\[1mm]
\ds{a_{2}\pars{x}} & \ds{=} & \ds{-1}
\end{array}\right.
\end{align}

For instance,
\begin{align}
\bbox[#ffd,10px,border:1px dotted navy]{\sin\pars{4x}} &\ = -\sin\pars{2x} + 2\sin\pars{3x}\cos\pars{x}
\\[5mm] & =
-\sin\pars{2x} +
2\bracks{-\sin\pars{x} + 2\sin\pars{2x}\cos\pars{x}}\cos\pars{x}
\\[5mm] & =
\bracks{4\cos^{2}\pars{x} - 1}\bracks{2\sin\pars{x}\cos\pars{x}} -
2\sin\pars{x}\cos\pars{x}
\\[5mm] & =\ \bbox[#ffd,10px,border:1px dotted navy]{%
8\sin\pars{x}\cos^{3}\pars{x} - 4\sin\pars{x}\cos\pars{x}}
\end{align}
A: The angle addition formulas for sine and cosine are nicely captured by this matrix formula
$$
\left( \begin{array}{c} s_{n} \\ c_{n}\end{array} \right) =
\left( \begin{array}{r} c & s \\ -s & c\end{array} \right) 
\left( \begin{array}{c} s_{n-1} \\ c_{n-1}\end{array} \right)
$$
where $s_k = \sin(kx)$, $c_k = \cos(kx)$, $s= s_1$ and $c=c_1$.
Playing this recursion backwards, we arrive at the formula
$$
\left( \begin{array}{c} s_{n} \\ c_{n}\end{array} \right) =
\left( \begin{array}{r} c & s \\ -s & c\end{array} \right)^n
\left( \begin{array}{c} s_{0} \\ c_{0}\end{array} \right)
$$
Note that $s_0=0$ and $c_0=1$.
It's worth pointing out that that
$$
\left( \begin{array}{r} c & s \\ -s & c\end{array} \right) =
c \left( \begin{array}{r} 1 & 0 \\ 0 & 1\end{array} \right) +
s \left( \begin{array}{r} 0 & 1 \\ -1 & 0\end{array} \right)
$$
and that
$$
\left( \begin{array}{r} 0 & 1 \\ -1 & 0 \end{array}\right)^2 =  
\left( \begin{array}{r} -1 & 0 \\ 0 & -1\end{array} \right)
$$
