# 在计算的时候尽可能将除法转换为乘法：乘法比除法更方便计算

## 二、解析

\begin{aligned} \cos x – 1 & = \frac{- x ^{2}}{2} + o(x ^{3}) \\ 1 – \sin x & = 1 – x + \frac{x ^{3}}{6} + o(x ^{3}) \end{aligned}

\begin{aligned} & \frac{\cos x – 1}{1 – \sin x} = a x + b x ^{2} + c x ^{3} + o(x ^{3}) \\ \\ \Rightarrow & \textcolor{magenta}{\cos x – 1} = \left[ a x + b x ^{2} + c x ^{3} + o(x ^{3}) \right] \cdot \left[ \textcolor{tan}{1 – \sin x} \right] \\ \\ \Rightarrow & \textcolor{magenta}{ \frac{- x ^{2}}{2} + o(x ^{3}) } = \left[ a x + b x ^{2} + c x ^{3} + o(x ^{3}) \right] \cdot \left[ \textcolor{tan}{ 1 – x + \frac{x ^{3}}{6} + o(x ^{3}) } \right] \\ \\ \Rightarrow & \frac{- x ^{2}}{2} = \left[ \textcolor{white}{\colorbox{green}{a}} x + \textcolor{white}{\colorbox{blue}{b}} x ^{2} + \textcolor{white}{\colorbox{red}{c}} x ^{3} \right] \cdot \left[ 1 – x + \frac{x ^{3}}{6} \right] \\ \\ \Rightarrow & \frac{-x ^{2}}{2} = \textcolor{white}{\colorbox{green}{a}} x + (\textcolor{white}{\colorbox{blue}{b}} – \textcolor{white}{\colorbox{green}{a}} ) x ^{2} + (\textcolor{white}{\colorbox{red}{c}} – \textcolor{white}{\colorbox{blue}{b}} ) x ^{3} \\ \\ \Rightarrow & 0 \cdot x + \frac{-1}{2} \cdot x ^{2} + 0 \cdot x ^{3} = \textcolor{white}{\colorbox{green}{a}} x + (\textcolor{white}{\colorbox{blue}{b}} – \textcolor{white}{\colorbox{green}{a}} ) x ^{2} + (\textcolor{white}{\colorbox{red}{c}} – \textcolor{white}{\colorbox{blue}{b}} ) x ^{3} \\ \\ \Rightarrow & \begin{cases} \textcolor{white}{\colorbox{green}{a}} = 0 \\ \textcolor{white}{\colorbox{blue}{b}} – \textcolor{white}{\colorbox{green}{a}} = \frac{-1}{2} \\ \textcolor{white}{\colorbox{red}{c}} – \textcolor{white}{\colorbox{blue}{b}} = 0 \end{cases} \\ \\ \Rightarrow & \textcolor{springgreen}{\boldsymbol{ \begin{cases} a = 0 \\ b = \frac{-1}{2} \\ c = \frac{1}{2} \end{cases} }} \end{aligned}