split a multiline bmatrix

Question

I was attempting to split up a 2 row bmatrix in two lines but was not successful. Can you help me split up the matrix. This is the effect that I wanted to get:

Here is my code:

\documentclass[11pt]{book}
\usepackage{amsmath,amsfonts,amssymb,amsthm, bm}

\begin{document}
\begin{equation}\label{theta}
\begin{split}
\begin{bmatrix}
x\\
y
\end{bmatrix} & =
\begin{bmatrix}
         \left(L_{1} + \delta r_{3v} + r_{3v} + \Delta r_{3}^{\circ}\right)\cos(\theta_{1}+ \delta\theta_{1})\\
         \left(L_{1} + \delta r_{3v} + r_{3v} + \Delta r_{3}^{\circ}\right)\sin(\theta_{1}+ \delta\theta_{1})
        \end{bmatrix}\\
&\qquad\qquad \begin{bmatrix}+(\delta r_{2v} + r_{2v} + \Delta r_{3}^{\circ})\sin(\theta_{1} + \delta\theta_{1})\\
- (\delta r_{2v} + r_{2v} + \Delta r_{3}^{\circ})\cos(\theta_{1} + \delta\theta_{1})
 \end{bmatrix}
 \end{split}
 \end{equation}
 \end{document} 

Answer 1

I wouldn't push down the left-hand side, so I provide both renderings.

\documentclass[11pt]{book}
\usepackage{amsmath,amsfonts,amssymb,amsthm, bm}

\begin{document}

\begin{equation}\label{theta}
\begin{bmatrix}
x\\
y
\end{bmatrix} =
\!\begin{aligned}
&
\left[\begin{matrix}
  L_{1} + \delta r_{3v} + r_{3v} + \Delta r_{3}^{\circ})\cos(\theta_{1}+ \delta\theta_{1})\\
  L_{1} + \delta r_{3v} + r_{3v} + \Delta r_{3}^{\circ})\sin(\theta_{1}+ \delta\theta_{1})
\end{matrix}\right.\\
&\qquad\qquad
\left.\begin{matrix}
  {}+(\delta r_{2v} + r_{2v} + \Delta r_{3}^{\circ})\sin(\theta_{1} + \delta\theta_{1})\\
  {}-(\delta r_{2v} + r_{2v} + \Delta r_{3}^{\circ})\cos(\theta_{1} + \delta\theta_{1})
\end{matrix}\right]
\end{aligned}
\end{equation}

\begin{equation}\label{theta-alt}
\begin{aligned}
\begin{bmatrix}
x\\
y
\end{bmatrix} &=
\left[\begin{matrix}
  L_{1} + \delta r_{3v} + r_{3v} + \Delta r_{3}^{\circ})\cos(\theta_{1}+ \delta\theta_{1})\\
  L_{1} + \delta r_{3v} + r_{3v} + \Delta r_{3}^{\circ})\sin(\theta_{1}+ \delta\theta_{1})
\end{matrix}\right.\\
&\qquad\qquad
\left.\begin{matrix}
  {}+(\delta r_{2v} + r_{2v} + \Delta r_{3}^{\circ})\sin(\theta_{1} + \delta\theta_{1})\\
  {}-(\delta r_{2v} + r_{2v} + \Delta r_{3}^{\circ})\cos(\theta_{1} + \delta\theta_{1})
\end{matrix}\right]
\end{aligned}
\end{equation}

\end{document} 

Note that I removed the superfluous \left( and \right). Note also the additional {} groups for making the plus and minus into binary operation symbols.

I would consider leaving the plus and minus in the top lines, so to make clear that the formula has not ended yet.

Answer 2

I'd rather use a placeholder such as \omega for \theta_{1}+ \delta\theta_{1} (change the name according to your needs) and explain it afterwards; in this way, your expression fits in one line, which is a better option here:

\documentclass[11pt]{book}
\usepackage{amsmath,amsfonts,amssymb,amsthm, bm}

\begin{document}
\begin{equation}\label{theta}
\begin{bmatrix}
x\\
y
\end{bmatrix} =
\begin{bmatrix}
         (L_{1} + \delta r_{3v} + r_{3v} + \Delta r_{3}^{\circ} )\cos\omega + ( \delta r_{2v} + r_{2v} + \Delta r_{3}^{\circ})\sin \omega \\
         (L_{1} + \delta r_{3v} + r_{3v} + \Delta r_{3}^{\circ} )\sin\omega - (\delta r_{2v} + r_{2v} + \Delta r_{3}^{\circ})\cos \omega
 \end{bmatrix},
 \end{equation}
where $\omega=\theta_{1}+ \delta\theta_{1}$.
\end{document} 

I also suppressed some unnecessary \left, \right pairs.