公式：已知 tan 的值，求 sin 和 cos 的值

一、前言

已知 $\tan \frac{x}{2}$ $=$ $t$, 则：

$$
\sin x = ?
$$

$$
\cos x = ?
$$

$$
\tan x = ?
$$

二、正文

分析可知，由于已知的是 $\frac{x}{2}$ 的 $\tan$ 值，要求解的确实 $x$ 的 $\sin$, $\cos$ 和 $\tan$ 值，所以，一定会用到三角函数的二倍角公式，即：

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

于是可知，想要求解出 $\sin$ 就需要先求解出 $\cos$, 至于 $\tan$ 由于和 $\sin$ 与 $\cos$ 都没有关系，所以可以在任意步骤完成求解。

$\cos x$

首先：

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

接下来，为了简化书写，可以令 $\cos \frac{x}{2}$ $=$ $\frac{x}{2}$, 于是：

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

根据前面的代换 “$\cos \frac{x}{2}$ $=$ $\frac{x}{2}$” 可知：

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

于是：

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

即：

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

$\sin x$

首先：

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

接下来，为了简化书写，可以令 $\sin \frac{x}{2}$ $=$ $\frac{x}{2}$, 于是：

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

根据前面的代换 “$\sin \frac{x}{2}$ $=$ $\frac{x}{2}$” 可知：

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

于是：

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

即：

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

$\tan x$

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

即：

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

结论

综上可知：

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

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