![方程式排列不正确](https://linux22.com/image/416735/%E6%96%B9%E7%A8%8B%E5%BC%8F%E6%8E%92%E5%88%97%E4%B8%8D%E6%AD%A3%E7%A1%AE.png)
问题:
我怎样才能保持下一页方程的连续性?第二个问题,我的一个方程太长了,我怎样才能打破它?
梅威瑟:
\documentclass[12pt,fleqn]{beamer}
\usetheme{AnnArbor}
\usepackage{setspace}
\usepackage{amsmath}
\usepackage[utf8]{inputenc}
%\setstretch{1.0}
\usecolortheme{beaver}
\usefonttheme{professionalfonts} % using non standard fonts for beamer
\usefonttheme{serif} % default family is serif
\addtobeamertemplate{frametitle}{}{\vspace{-0.4em}} % decrease
\makeatletter
\newcommand{\Pause}[1][]{\unless\ifmeasuring@\relax
\pause[#1]%
\fi}
\makeatother
\title[pqr lmn] %optional
{Chapter}
\subtitle{Lecture - 0}
\author[X. Y. Z] % (optional, for multiple authors)
{x.~y.~z \\ abc \\ def}
%\date[\today] % (optional)
\begin{document}
\frame{\titlepage}
\begin{frame}
%\setstretch{1.0}
\textbf{Find nth order derivative of} $\boldsymbol{\cos x \cos 2x \cos 3x}$
\begin{align*}
y&=\cos x \cos 2x \cos 3x\\[8pt]
&=\dfrac{1}{2}\cos x \left(2\,\cos 3x \cos 2x\right)\\[12pt]
&=\dfrac{1}{2}\,\cos x \,\left(\cos 5x + \cos x\right)\\[12pt]
&=\dfrac{1}{4}\,\left[2 \cos 5x \cos x + 2\cos^2 x\right]\\[12pt]
&=\dfrac{1}{4}\,\left[\cos 6x + \cos 4x + 1 + \cos 2x\right]
\end{align*}
\end{frame}
\begin{frame}
%\setstretch{1.0}
$$\therefore\;\;y_n=\dfrac{1}{4}\,\left[\left\{\cos 6x\right\}_n + \left\{\cos 4x\right\}_n + \left\{1\right\}_n + \left\{\cos 2x\right\}_n\right]$$
$$=\dfrac{1}{4}\,\left[6^n\,\left\{\cos 6x+\dfrac{n\pi}{2}\right\} + 4^n\,\left\{\cos 4x+\dfrac{n\pi}{2}\right\} + 0 + 2^n\,\left\{\cos 2x+\dfrac{n\pi}{2}\right\}\right]$$
\end{frame}
\end{document}