数学-三角関数-加法定理の証明

# 加法定理の証明

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[余弦定理の証明](2c5183058e4ffdb32b9a)

## cos(α-β)の証明

![kahouteiri-3.gif](https://qiita-image-store.s3.amazonaws.com/0/4641/635192ab-1722-4efc-b6ab-67c5f5cbf10e.gif)

```math
\\
点P(\cos \alpha, \sin \alpha)、Q(\cos \beta, \sin \beta) とする。\\
```

```math
\\
\begin{aligned}
余弦定理より、\\
PQ^2 &= OP^2 + OQ^2 - 2OP \cdot OQ \cdot \cos(\alpha - \beta) \\
&= 1^2 + 1^2 - 2 \cdot 1 \cdot 1 \cdot \cos(\alpha - \beta) \\
&= 2 - 2 \cos(\alpha - \beta) ・・・① \\
\end{aligned}\\
```

```math
\\
\begin{aligned}
線分の長さから、\\
PQ^2 &= (\cos \beta - \cos \alpha)^2 + (\sin \beta - \sin \alpha)^2 \\
     &= (\cos^2 \alpha - 2 \cos \alpha \cos \beta + \cos^2 \beta) + (\sin^2 \alpha - 2 \sin \alpha \sin \beta + \sin^2 \beta)  \\
     &= (\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) - 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta)\\
     &= 2 - 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta)・・・②\\
\end{aligned}\\
```

```math
\\
\begin{aligned}
①②より\\
&\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\
\end{aligned}
```

## cos(α+β)の証明

```math
\\
\begin{array}{l}
\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\
\\
上記に、\beta = -\beta, （\sin(- \beta) = - \sin \beta,　\cos(- \beta) = \cos \beta） を代入 \\
\\
\cos(\alpha - (-\beta)) = \cos \alpha \cos (-\beta) + \sin \alpha \sin (-\beta) \\
\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\
\end{array}
```

## sin(α-β)の証明

```math
\\
\begin{array}{l}
\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\
\\
上記に、 \beta = \beta + \frac{\pi}{2} （\sin(\beta + \frac{\pi}{2}) = \cos \beta,　\cos(\beta + \frac{\pi}{2}) = - \sin \beta） を代入 \\
\\
左辺に\beta = \beta + \frac{\pi}{2}を代入\\
\cos(\alpha - \beta) = \cos(\alpha - (\beta + \frac{\pi}{2})) = \cos((\alpha - \beta) - \frac{\pi}{2}) = \cos(\frac{\pi}{2} - (\alpha - \beta)) = \sin(\alpha - \beta)\\
\\
右辺に\beta = \beta + \frac{\pi}{2}を代入\\
\cos \alpha \cos \beta + \sin \alpha \sin \beta \\
　 = \cos \alpha \cos (\beta + \frac{\pi}{2}) + \sin \alpha \sin (\beta + \frac{\pi}{2}) \\
　 = - \cos \alpha \sin \beta + \sin \alpha \cos \beta \\
　 = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\
\\
\therefore \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta\\
\end{array}
```


## sin(α+β)の証明

```math
\\
\begin{array}{l}
\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\
\\
上記に、\beta = -\beta （\sin(- \beta) = - \sin \beta,　\cos(- \beta) = \cos = \beta） を代入 \\
\\
\sin(\alpha - (-\beta)) = \sin \alpha \cos (-\beta) - \cos \alpha \sin (-\beta) \\
\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\
\end{array}
```


## tan(α+β)の証明

```math
\\
\begin{array}{ll}
\tan(\alpha + \beta) &= \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)}\\
&= \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} 　※\cos \alpha \cos \beta で割る\\
&= \frac{\frac{\sin \alpha \cos \beta}{\cos \alpha \cos \beta} + \frac{\cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \cos \beta}{\cos \alpha \cos \beta} - \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}}　※\frac{\sin \theta}{\cos \theta} = \tan \theta\\
&= \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \\
\end{array}
\\
```

## tan(α-β)の証明

```math
\\
\begin{array}{ll}
\tan(\alpha - \beta) &= \frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)}\\
&= \frac{\sin \alpha \cos \beta - \cos \alpha \sin \beta}{\cos \alpha \cos \beta + \sin \alpha \sin \beta} 　※\cos \alpha \cos \beta で割る\\
&= \frac{\frac{\sin \alpha \cos \beta}{\cos \alpha \cos \beta} - \frac{\cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \cos \beta}{\cos \alpha \cos \beta} + \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}}　※\frac{\sin \theta}{\cos \theta} = \tan \theta\\
&= \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \\
\end{array}
```