# 公式：已知 tan 的值，求 sin 和 cos 的值

## 一、前言

$$\sin x = ?$$

$$\cos x = ?$$

$$\tan x = ?$$

## 二、正文

\begin{aligned} \sin 2 \alpha & = 2 \sin \alpha \cos \alpha \\ \cos 2 \alpha & = 2 \cos^{2} \alpha – 1 \\ \tan 2 \alpha & = \frac{2 \tan \alpha}{1 – \tan^{2} \alpha} \end{aligned}

### $\cos x$

\begin{aligned} \tan \frac{x}{2} & = t \\ \\ & \Rightarrow \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = t \\ \\ & \Rightarrow \textcolor{orangered}{\frac{\sqrt{1 – \cos^{2} \frac{x}{2}}}{\cos \frac{x}{2}} = t} \end{aligned}

\begin{aligned} \textcolor{orangered}{\frac{\sqrt{1 – \cos^{2} \frac{x}{2}}}{\cos \frac{x}{2}} = t} \\ \\ & \Rightarrow \frac{\sqrt{1 – (\frac{x}{2})^{2}}}{\frac{x}{2}} = t \\ \\ & \Rightarrow \frac{1 – (\frac{x}{2})^{2}}{(\frac{x}{2})^{2}} = t^{2} \\ \\ & \Rightarrow 1 – (\frac{x}{2})^{2} = t^{2} (\frac{x}{2})^{2} \\ \\ & \Rightarrow (t^{2} + 1) \cdot (\frac{x}{2})^{2} = 1 \\ \\ & \Rightarrow (\frac{x}{2})^{2} = \frac{1}{1 + t^{2}} \\ \\ & \Rightarrow \textcolor{yellow}{ \frac{x}{2} = \frac{1}{\sqrt{1+t^{2}}} } \end{aligned}

$$\textcolor{yellow}{ \cos \frac{x}{2} = \frac{1}{\sqrt{1+t^{2}}} }$$

\begin{aligned} \cos x & = 2 \textcolor{yellow}{\cos^{2} \frac{x}{2} } – 1 \\ \\ & = 2 \cdot (\frac{1}{\sqrt{1+t^{2}}})^{2} – 1 \\ \\ & = \frac{2}{1 + t^{2}} – 1 \\ \\ & = \textcolor{springgreen}{ \frac{1 – t^{2}}{1 + t^{2}} } \end{aligned}

$$\textcolor{green}{ \boldsymbol{ \cos x = \frac{1 – t^{2}}{1 + t^{2}} } }$$

### $\sin x$

\begin{aligned} \tan \frac{x}{2} = t \\ \\ & \Rightarrow \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = t \\ \\ & \Rightarrow \textcolor{orangered}{ \frac{\sin \frac{x}{2}}{\sqrt{1 – \sin^{2} \frac{x}{2}}} = t } \end{aligned}

\begin{aligned} \textcolor{orangered}{ \frac{\sin \frac{x}{2}}{\sqrt{1 – \sin^{2} \frac{x}{2}}} = t } \\ \\ & \Rightarrow \frac{\frac{x}{2}}{\sqrt{1 – (\frac{x}{2})^{2}}} = t \\ \\ & \Rightarrow \frac{(\frac{x}{2})^{2}}{1 – (\frac{x}{2})^{2}} = t^{2} \\ \\ & \Rightarrow (\frac{x}{2})^{2} = t^{2} – t^{2} (\frac{x}{2})^{2} \\ \\ & \Rightarrow (1 + t^{2}) \cdot (\frac{x}{2})^{2} = t^{2} \\ \\ & \Rightarrow \textcolor{yellow}{ \frac{x}{2} = \frac{t}{\sqrt{1 + t^{2}}} } \end{aligned}

$$\textcolor{yellow}{ \sin \frac{x}{2} = \frac{t}{\sqrt{1 + t^{2}}} }$$

\begin{aligned} \sin x & = 2 \textcolor{yellow}{ \sin \frac{x}{2} } \cdot \textcolor{yellow}{ \cos \frac{x}{2} } \\ \\ & = 2 \cdot \frac{t}{\sqrt{1 + t^{2}}} \cdot \frac{1}{\sqrt{1+t^{2}}} \\ \\ & = \textcolor{springgreen}{ \frac{2t}{1 + t^{2}} } \end{aligned}

$$\textcolor{green}{ \boldsymbol{ \sin x = \frac{2t}{1 + t^{2}} } }$$

### $\tan x$

\begin{aligned} \tan x & = \frac{2 \tan \frac{x}{2}}{1 – \tan ^{2} \frac{x}{2}} \\ \\ & = \frac{2t}{1 – t^{2}} \end{aligned}

$$\textcolor{green}{ \boldsymbol{ \tan x = \frac{2t}{1 – t^{2}} } }$$

### 结论

\textcolor{green}{ \begin{aligned} \boldsymbol{ \sin x } & = \boldsymbol{\frac{2t}{1 + t^{2}} } \\ \\ \boldsymbol{ \cos x } & = \boldsymbol{ \frac{1 – t^{2}}{1 + t^{2}} } \\ \\ \boldsymbol{ \tan x } & = \boldsymbol{ \frac{2t}{1 – t^{2}} } \end{aligned} }