Staggered levels of alignment ; align to parts of previous line that weren't aligned with the lines before it

Question

How can I align the maths equation as I've aligned it in the source?

\begin{align}
         &\sin(A+B)=\sin A\cos B + \sin B\cos A \\
         &\sin(A-B)=\sin A\cos B - \sin B\cos A
%
  \intertext{It's possible to eliminate the $\sin B\cos A$ terms by summing the two
  equations, it will then be possible to rearrange for
  $\sin A\cos B$:}
%
(2)+(3): &\sin(A+B) + \sin(A-B)&=2\sin A\cos B \nonumber\\
                 2\sin A\cos B &=\sin(A+B) + \sin(A-B) \nonumber
\end{align}

Here's what it currently looks like, with annotations as to how I'd like it to look:

I don't really care about it being centred, I think it just looks uncomfortably left-aligned as it currently is.

I plan to add more lines of display maths below the current last one, which I'd also like to align with the equal sign of that first unnumbered line.

Answer 1

Nested aligned, with less code and the correct numbers without guessing at them.

However, the alignment is not at all necessary, particularly in the last line. I add an alternative solution. The reverse equality doesn't seem useful, though.

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\section{How you want it}

\begin{align}
         &\sin(A+B)=\sin A\cos B + \sin B\cos A \label{sum}\\
         &\sin(A-B)=\sin A\cos B - \sin B\cos A \label{diff}
\intertext{It's possible to eliminate the $\sin B\cos A$ terms by summing the two
  equations, it will then be possible to rearrange for
  $\sin A\cos B$:}
\eqref{sum}+\eqref{diff}{:}\qquad
      &\begin{aligned}[t]
    \sin(A+B) + \sin(A-B)&=2\sin A\cos B \\
    2\sin A\cos B &=\sin(A+B) + \sin(A-B)
   \end{aligned}
  \notag
\end{align}

\section{How I would do it}

\begin{gather}
  \sin(A+B)=\sin A\cos B + \sin B\cos A \label{sum+}\\
  \sin(A-B)=\sin A\cos B - \sin B\cos A \label{diff+}
\end{gather}
It's possible to eliminate the $\sin B\cos A$ terms by summing the two
equations, it will then be possible to rearrange for $\sin A\cos B$:
\begin{equation*}
\makebox[0pt][r]{$[\eqref{sum+}+\eqref{diff+}]$\qquad}
\begin{gathered}
  \sin(A+B) + \sin(A-B)=2\sin A\cos B \\
  2\sin A\cos B =\sin(A+B) + \sin(A-B)
\end{gathered}
\end{equation*}

\end{document} 

Answer 2

Use a nested aligned:

\documentclass{article}
\usepackage{mathtools, nccmath}

\begin{document}

\begin{align}
         &\sin(A+B)=\sin A\cos B + \sin B\cos A \\
         &\sin(A-B)=\sin A\cos B - \sin B\cos A
%
  \intertext{It's possible to eliminate the $\sin B\cos A$ terms by summing the two
  equations, it will then be possible to rearrange for
  $\sin A\cos B$:}
%
(2)+(3): \qquad\notag&\smash{\begin{aligned}[t]\sin(A+B) + \sin(A-B)&=2\sin A\cos B \\
                 2\sin A\cos B &=\sin(A+B) + \sin(A-B)
                 \end{aligned}}\\[-0.05ex]%
\end{align}

\end{document}