三角関数計算の基礎

■関係式

1.   $sin\left(-\theta \right)=-sin\left(\theta \right)$ ， ${\displaystyle cos\left(-\theta \right)=cos\left(\theta \right)}$ ， ${\displaystyle tan\left(-\theta \right)=-tan\theta }$   ⇒導出
2.   $sin\left(90°-\theta \right)=cos\theta$ ， ${\displaystyle cos\left(90°-\theta \right)=sin\theta }$ ， ${\displaystyle tan\left(90°-\theta \right)=\frac{1}{tan\theta }}$   ⇒導出（ここも参照）
3.   $sin\left(90°+\theta \right)=cos\theta$ ， ${\displaystyle cos\left(90°+\theta \right)=-sin\theta }$ ， ${\displaystyle tan\left(90°+\theta \right)=-\frac{1}{tan\theta }}$   ⇒導出
4.   $sin\left(180°-\theta \right)=sin\theta$ ， ${\displaystyle cos\left(180°-\theta \right)=-cos\theta }$ ， ${\displaystyle tan\left(180°-\theta \right)=-tan\theta }$   ⇒導出
5.   $sin\left(180°+\theta \right)=-sin\theta$ ， ${\displaystyle cos\left(180°+\theta \right)=-cos\theta }$ ， ${\displaystyle tan\left(180°+\theta \right)=tan\theta }$   ⇒導出

■sinとcosのグラフの関係

　sinとcosの関係式を理解するのに役に立つグラフである．

■導出

1.三角関数の定義より

${\displaystyle \{\begin{array}{l}sin\theta =y\\ cos\theta =x\\ tan\theta =\frac{y}{x}\end{array}}$    ，  ${\displaystyle \{\begin{array}{l}sin\left(-\theta \right)=-y\\ cos\left(-\theta \right)=x\\ tan\left(-\theta \right)=\frac{-y}{x}\end{array}}$

よって

$sin\left(-\theta \right)=-sin\left(\theta \right)$ ， $cos\left(-\theta \right)=cos\left(\theta \right)$ , ${\displaystyle tan\left(-\theta \right)=-tan\theta }$

2.三角関数の定義より

${\displaystyle \{\begin{array}{l}sin\theta =y\\ cos\theta =x\\ tan\theta =\frac{y}{x}\end{array}}$    ，  ${\displaystyle \{\begin{array}{l}sin\left(90°-\theta \right)=x\\ cos\left(90°-\theta \right)=y\\ tan\left(90°-\theta \right)=\frac{x}{y}\end{array}}$

よって

$sin\left(90°-\theta \right)=cos\theta$ ， $cos\left(90°-\theta \right)=sin\theta$ ， ${\displaystyle tan\left(90°-\theta \right)=\frac{1}{tan\theta }}$

3.三角関数の定義より

${\displaystyle \{\begin{array}{l}sin\theta =y\\ cos\theta =x\\ tan\theta =\frac{y}{x}\end{array}}$    ，  ${\displaystyle \{\begin{array}{l}sin\left(90°+\theta \right)=x\\ cos\left(90°+\theta \right)=-y\\ tan\left(90°+\theta \right)=\frac{x}{-y}\end{array}}$

よって

$sin\left(90°+\theta \right)=cos\theta$ ， $cos\left(90°+\theta \right)=-sin\theta$ ， ${\displaystyle tan\left(90°+\theta \right)=-\frac{1}{tan\theta }}$

4.三角関数の定義より

${\displaystyle \{\begin{array}{l}sin\theta =y\\ cos\theta =x\\ tan\theta =\frac{y}{x}\end{array}}$    ，  ${\displaystyle \{\begin{array}{l}sin\left(180°-\theta \right)=y\\ cos\left(180°-\theta \right)=-x\\ tan\left(180°-\theta \right)=\frac{y}{-x}\end{array}}$

よって

$sin\left(180°-\theta \right)=sin\theta$ ， $cos\left(180°-\theta \right)=-cos\theta$ ， ${\displaystyle tan\left(180°-\theta \right)=-tan\theta }$

5.三角関数の定義より

${\displaystyle \{\begin{array}{l}sin\theta =y\\ cos\theta =x\\ tan\theta =\frac{y}{x}\end{array}}$    ，  ${\displaystyle \{\begin{array}{l}sin\left(180°+\theta \right)=-y\\ cos\left(180°+\theta \right)=-x\\ tan\left(180°+\theta \right)=\frac{-y}{-x}\end{array}}$

よって

$sin\left(180°+\theta \right)=-sin\theta$ ， $cos\left(180°+\theta \right)=-cos\theta$ ， ${\displaystyle tan\left(180°+\theta \right)=tan\theta }$

 

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最終更新日：2026年2月10日