三角関数計算の基礎

# 三角関数計算の基礎

## ■関係式

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

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

よって，

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

2.三角関数の定義より，

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

よって，

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

3.三角関数の定義より，

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

よって，

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

4.三角関数の定義より，

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

よって，

$sin\left(180°-\theta \right)=sin\theta$$cos\left(180°-\theta \right)=-cos\theta$$tan\left(180°-\theta \right)=-tan\theta$

5.三角関数の定義より，

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

よって，

$sin\left(180°+\theta \right)=-sin\theta$$cos\left(180°+\theta \right)=-cos\theta$$tan\left(180°+\theta \right)=tan\theta$

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