三角関数計算の基礎

■関係式

$3$\sin $ - {\theta} $ =- \sin ${\theta}$$	$3$\cos $ - {\theta} $ =\cos ${\theta}$$	$3$\displaystyle{\tan $ - {\theta} $ =- \tan {\theta}}$	⇒導出
$3$\sin $ 90^\circ - {\theta} $ =\cos {\theta}$	$3$\cos $ 90^\circ - {\theta} $ =\sin {\theta}$	$3$\displaystyle{\tan $ 90^\circ - {\theta} $ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} \tan {\theta} \hspace{2}}}$	⇒導出 ⇒ここも参照
$3$\sin $ 90^\circ +{\theta} $ =\cos {\theta}$	$3$\cos $ 90^\circ +{\theta} $ =- \sin {\theta}$	$3$\displaystyle{\tan $ 90^\circ +{\theta} $ =- \frac{\hspace{2}1\hspace{2}}{\hspace{2} \tan {\theta} \hspace{2}}}$	⇒導出
$3$\sin $ 180^\circ - {\theta} $ =\sin {\theta}$	$3$\cos $ 180^\circ - {\theta} $ =- \cos {\theta}$	$3$\displaystyle{\tan $ 180^\circ - {\theta} $ =- \tan {\theta}}$	⇒導出
$3$\sin $ 180^\circ +{\theta} $ =- \sin {\theta}$	$3$\cos $ 180^\circ +{\theta} $ =- \cos {\theta}$	$3$\displaystyle{\tan $ 180^\circ +{\theta} $ =\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，

$3$\displaystyle{ \{\begin{array}{lll}\sin {\theta}=y& \vspace{6}\\ \cos {\theta}=x& \vspace{6}\\ \tan {\theta}=\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}& \vspace{6}\\ \end{array} }$ ， $3$\displaystyle{ \{\begin{array}{lll}\sin $ - {\theta} $ =- y& \vspace{6}\\ \cos $ - {\theta} $ =x& \vspace{6}\\ \tan $ - {\theta} $ =\frac{\hspace{2} - y \hspace{2}}{\hspace{2}x\hspace{2}}& \vspace{6}\\ \end{array} }$

よって，

$3$\sin $ - {\theta} $ =- \sin ${\theta}$$ ， $3$\cos $ - {\theta} $ =\cos ${\theta}$$ , $3$\displaystyle{\tan $ - {\theta} $ =- \tan {\theta}}$

2.三角関数の定義より，

$3$\displaystyle{ \{\begin{array}{lll}\sin {\theta}=y& \vspace{6}\\ \cos {\theta}=x& \vspace{6}\\ \tan {\theta}=\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}& \vspace{6}\\ \end{array} }$ ， $3$\displaystyle{ \{\begin{array}{lll}\sin $ 90^\circ - {\theta} $ =x& \vspace{6}\\ \cos $ 90^\circ - {\theta} $ =y& \vspace{6}\\ \tan $ 90^\circ - {\theta} $ =\frac{\hspace{2} x \hspace{2}}{\hspace{2}y\hspace{2}}& \vspace{6}\\ \end{array} }$

よって，

$3$\sin $ 90^\circ - {\theta} $ =\cos {\theta}$ ， $3$\cos $ 90^\circ - {\theta} $ =\sin {\theta}$ ， $3$\displaystyle{\tan $ 90^\circ - {\theta} $ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} \tan {\theta} \hspace{2}}}$

3.三角関数の定義より，

$3$\displaystyle{ \{\begin{array}{lll}\sin {\theta}=y& \vspace{6}\\ \cos {\theta}=x& \vspace{6}\\ \tan {\theta}=\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}& \vspace{6}\\ \end{array} }$ ， $3$\displaystyle{ \{\begin{array}{lll}\sin $ 90^\circ +{\theta} $ =x& \vspace{6}\\ \cos $ 90^\circ +{\theta} $ =- y& \vspace{6}\\ \tan $ 90^\circ +{\theta} $ =\frac{\hspace{2} x \hspace{2}}{\hspace{2} - y \hspace{2}}& \vspace{6}\\ \end{array} }$

よって，

$3$\sin $ 90^\circ +{\theta} $ =\cos {\theta}$ ， $3$\cos $ 90^\circ +{\theta} $ =- \sin {\theta}$ ， $3$\displaystyle{\tan $ 90^\circ +{\theta} $ =- \frac{\hspace{2}1\hspace{2}}{\hspace{2} \tan {\theta} \hspace{2}}}$

4.三角関数の定義より，

$3$\displaystyle{ \{\begin{array}{lll}\sin {\theta}=y& \vspace{6}\\ \cos {\theta}=x& \vspace{6}\\ \tan {\theta}=\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}& \vspace{6}\\ \end{array} }$ ， $3$\displaystyle{ \{\begin{array}{lll}\sin $ 180^\circ - {\theta} $ =y& \vspace{6}\\ \cos $ 180^\circ - {\theta} $ =- x& \vspace{6}\\ \tan $ 180^\circ - {\theta} $ =\frac{\hspace{2} y \hspace{2}}{\hspace{2} - x \hspace{2}}& \vspace{6}\\ \end{array} }$

よって，

$3$\sin $ 180^\circ - {\theta} $ =\sin {\theta}$ ， $3$\cos $ 180^\circ - {\theta} $ =- \cos {\theta}$ ， $3$\displaystyle{\tan $ 180^\circ - {\theta} $ =- \tan {\theta}}$

5.三角関数の定義より，

$3$\displaystyle{ \{\begin{array}{lll}\sin {\theta}=y& \vspace{6}\\ \cos {\theta}=x& \vspace{6}\\ \tan {\theta}=\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}& \vspace{6}\\ \end{array} }$ ， $3$\displaystyle{ \{\begin{array}{lll}\sin $ 180^\circ +{\theta} $ =- y& \vspace{6}\\ \cos $ 180^\circ +{\theta} $ =- x& \vspace{6}\\ \tan $ 180^\circ +{\theta} $ =\frac{\hspace{2} - y \hspace{2}}{\hspace{2} - x \hspace{2}}& \vspace{6}\\ \end{array} }$

よって，

$3$\sin $ 180^\circ +{\theta} $ =- \sin {\theta}$ ， $3$\cos $ 180^\circ +{\theta} $ =- \cos {\theta}$ ， $3$\displaystyle{\tan $ 180^\circ +{\theta} $ =\tan {\theta}}$

ホーム>>カテゴリー分類>>三角関数>>三角関数計算の基礎

最終更新日：2025年4月27日

$3$\sin \( - {\theta} \) =- \sin \({\theta}\)$	$3$\cos \( - {\theta} \) =\cos \({\theta}\)$	$3$\displaystyle{\tan \( - {\theta} \) =- \tan {\theta}}$	⇒導出
$3$\sin \( 90^\circ - {\theta} \) =\cos {\theta}$	$3$\cos \( 90^\circ - {\theta} \) =\sin {\theta}$	$3$\displaystyle{\tan \( 90^\circ - {\theta} \) =\frac{\hspace{2}1\hspace{2}}{\hspace{2} \tan {\theta} \hspace{2}}}$	⇒導出 ⇒ここも参照
$3$\sin \( 90^\circ +{\theta} \) =\cos {\theta}$	$3$\cos \( 90^\circ +{\theta} \) =- \sin {\theta}$	$3$\displaystyle{\tan \( 90^\circ +{\theta} \) =- \frac{\hspace{2}1\hspace{2}}{\hspace{2} \tan {\theta} \hspace{2}}}$	⇒導出
$3$\sin \( 180^\circ - {\theta} \) =\sin {\theta}$	$3$\cos \( 180^\circ - {\theta} \) =- \cos {\theta}$	$3$\displaystyle{\tan \( 180^\circ - {\theta} \) =- \tan {\theta}}$	⇒導出
$3$\sin \( 180^\circ +{\theta} \) =- \sin {\theta}$	$3$\cos \( 180^\circ +{\theta} \) =- \cos {\theta}$	$3$\displaystyle{\tan \( 180^\circ +{\theta} \) =\tan {\theta}}$	⇒導出