# Problem with my table on Latex

I am new to Latex. I built the following table and I have two problems. One is that raw number 7 in columns 1 and 2 has a width that is wider than the width of the other cells in the same columns. The second is that two equations in the lower part of the table (in column 3&4 in the merged raws from 7:9) are not centered. I have tried several things to correct the issues. I am now asking for help. How can these problems be corrected ? Thanks for any, and all, help.

Here is my Latex code, which has been made with use of OverLeaf:

\begin{table}[H]
\centering
\renewcommand{\arraystretch}{1.5}  % Adjust the height of the rows
\small  % Adjust the font size
\noindent
\setlength\tabcolsep{6pt}

% Calculate the maximum width for columns 1 and 2
\newlength{\maxwidth}
\settowidth{\maxwidth}{$\alpha_{2}<0, \, \alpha_{1}>0$}

\begin{tabular}{p{\maxwidth}|p{\maxwidth}|c|c}
angles sign & compare angles & $R$ & $z$ \\ \hline
$\alpha_{2}>0, \, \alpha_{1}>0$ & $\alpha_{2}>\alpha_{1}$ & \multirow{5}{*}{%
\begin{aligned} R_{2} &= R_{1}+\frac{1}{2}c_{1}\sin(\alpha_{1}) -\frac{1}{2}b_{1}\cos(\alpha_{1}) \\& +\frac{1}{2}c_{2}\sin(\alpha_{2}) +\frac{1}{2}b_{2}\cos(\alpha_{2}) \end{aligned}} & \multirow{5}{*}{%
\begin{aligned} z_{2} &= z_{1}+\frac{1}{2}c_{1}\cos(\alpha_{1}) -\frac{1}{2}b_{1}\sin(\alpha_{1}) \\ & +\frac{1}{2}c_{2}\cos(\alpha_{2}) +\frac{1}{2}b_{2}\sin(\alpha_{2}) \end{aligned}} \\ \cline{1-2}\cline{1-2}
$\alpha_{2}>0, \, \alpha_{1}<0$ & $\alpha_{2}>\alpha_{1}$ & & \\ \cline{1-2}\cline{1-2}
$\alpha_{2}<0, \, \alpha_{1}<0$ &  $\alpha_{2}>\alpha_{1}$ & & \\ \cline{1-2}\cline{1-2}
$\alpha_{2}>0, \, \alpha_{1}>0$ & $\alpha_{2}=\alpha_{1}$ & & \\ \cline{1-2}\cline{1-2}
$\alpha_{2}<0, \, \alpha_{1}<0$ & $\alpha_{2}=\alpha_{1}$ & & \\ \cline{1-4}
$\alpha_{2}<0, \, \alpha_{1}>0$ &  $\alpha_{2}<\alpha_{1}$ &
\begin{aligned} R_{2} &= R_{1}+\frac{1}{2}c_{1}\sin(\alpha_{1}) -\frac{1}{2}b_{1}\cos(\alpha_{1})\\ & +\frac{1}{2}c_{2}\sin(\alpha_{2})+\frac{1}{2}b_{2}\cos(\alpha_{2}) \end{aligned}  &
\begin{aligned} z_{2} &= z_{1}+\frac{1}{2}c_{1}\cos(\alpha_{1})+\frac{1}{2}b_{1}\sin(\alpha_{1})\\ &+\frac{1}{2}c_{2}\cos(\alpha_{2}) -\frac{1}{2}b_{2}\sin(\alpha_{2}) \end{aligned} \\ \cline{1-2}\cline{1-2}
$\alpha_{2}>0, \, \alpha_{1}>0$ & $\alpha_{2}<\alpha_{1}$ & & \\ \cline{1-2}\cline{1-2}
$\alpha_{2}<0, \, \alpha_{1}<0$&  $\alpha_{2}<\alpha_{1}$ & & \\ \hline
\end{tabular}
\renewcommand{\arraystretch}{1}  % Reset the array stretch for the rest of the document
\end{table}


I'd exploit the fact that the objects in the first two columns have the same widths. For the split formulas I'd exploit their symmetries and flush them right.

\documentclass{article}
\usepackage{amsmath}
\usepackage{booktabs}

\begin{document}

\begin{table}
\centering\small

%\setlength\tabcolsep{6pt}

\begin{tabular}{@{}cccc@{}}
\toprule
\multicolumn{2}{@{}c}{Angles} & $R$ & $z$ \\
\cmidrule(r){1-2}
sign & size \\
\midrule
\begin{tabular}{@{}c@{}}
$\alpha_{2}>0, \, \alpha_{1}>0$ \\
$\alpha_{2}>0, \, \alpha_{1}<0$ \\
$\alpha_{2}<0, \, \alpha_{1}<0$ \\
$\alpha_{2}>0, \, \alpha_{1}>0$ \\
$\alpha_{2}<0, \, \alpha_{1}<0$
\end{tabular} &
\begin{tabular}{@{}c@{}}
$\alpha_{2}>\alpha_{1}$ \\
$\alpha_{2}>\alpha_{1}$ \\
$\alpha_{2}>\alpha_{1}$ \\
$\alpha_{2}=\alpha_{1}$ \\
$\alpha_{2}=\alpha_{1}$
\end{tabular} &
\begin{aligned} R_{2} = R_{1}+\frac{1}{2}c_{1}\sin(\alpha_{1})-\frac{1}{2}b_{1}\cos(\alpha_{1}) \\[0.5ex] +\frac{1}{2}c_{2}\sin(\alpha_{2})+\frac{1}{2}b_{2}\cos(\alpha_{2}) \end{aligned} &
\begin{aligned} z_{2} = z_{1}+\frac{1}{2}c_{1}\cos(\alpha_{1})-\frac{1}{2}b_{1}\sin(\alpha_{1}) \\[0.5ex] +\frac{1}{2}c_{2}\cos(\alpha_{2}) +\frac{1}{2}b_{2}\sin(\alpha_{2}) \end{aligned} \\
\midrule
\begin{tabular}{@{}c@{}}
$\alpha_{2}<0, \, \alpha_{1}>0$ \\
$\alpha_{2}>0, \, \alpha_{1}>0$ \\
$\alpha_{2}<0, \, \alpha_{1}<0$
\end{tabular} &
\begin{tabular}{@{}c@{}}
$\alpha_{2}<\alpha_{1}$ \\
$\alpha_{2}<\alpha_{1}$ \\
$\alpha_{2}<\alpha_{1}$
\end{tabular} &
\begin{aligned} R_{2} = R_{1}+\frac{1}{2}c_{1}\sin(\alpha_{1})-\frac{1}{2}b_{1}\cos(\alpha_{1}) \\[0.5ex] +\frac{1}{2}c_{2}\sin(\alpha_{2})+\frac{1}{2}b_{2}\cos(\alpha_{2}) \end{aligned}  &
\begin{aligned} z_{2} = z_{1}+\frac{1}{2}c_{1}\cos(\alpha_{1})+\frac{1}{2}b_{1}\sin(\alpha_{1}) \\[0.5ex] +\frac{1}{2}c_{2}\cos(\alpha_{2})-\frac{1}{2}b_{2}\sin(\alpha_{2}) \end{aligned} \\