一般角の三角関数
【一般角の三角関数】
このとき、一般角\(\theta\)を次のように定義する。
\(\displaystyle \sin\theta=\frac{y}{r}\)
\(\displaystyle \cos\theta=\frac{x}{r}\)
\(\displaystyle \tan\theta=\frac{y}{x}\)
ただし、\(\tan\theta\)は\(x\neq0\)
これらを\(\theta\)の三角関数という。
【三角関数の値域】
原点を中心とする半径\(1\)の円を単位円という。単位円の三角関数は、
\(\sin\theta=y\)
\(\cos\theta=x\)
\(\displaystyle \tan\theta=\frac{y}{x}\)
単位円における三角関数の値域は、
\(-1\leqq\sin\theta\leqq1\)
\(-1\leqq\cos\theta\leqq1\)
\(\tan\theta\)は全ての実数値
【例題】次の\(\theta\)について、\(\sin\theta,\cos\theta,\tan\theta\)の値を求めなさい。
(1)\(\displaystyle \frac{\pi}{4}\)
\(\displaystyle \sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}\)
\(\displaystyle \cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\)
\(\displaystyle \tan\frac{\pi}{4}=1\)
(2)\(\displaystyle \frac{\pi}{3}\)
\(\displaystyle \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\)
\(\displaystyle \cos\frac{\pi}{3}=\frac{1}{2}\)
\(\displaystyle \tan\frac{\pi}{3}=\sqrt{3}\)
(3)\(\displaystyle \frac{2}{3}\pi\)
\(\displaystyle \sin\frac{2}{3}\pi=\frac{\sqrt{3}}{2}\)
\(\displaystyle \cos\frac{2}{3}\pi=-\frac{1}{2}\)
\(\displaystyle \tan\frac{2}{3}\pi=-\sqrt{3}\)
(4)\(\displaystyle \frac{7}{6}\pi\)
\(\displaystyle \sin\frac{7}{6}\pi=-\frac{1}{2}\)
\(\displaystyle \cos\frac{7}{6}\pi=-\frac{\sqrt{3}}{2}\)
\(\displaystyle \tan\frac{7}{6}\pi=\frac{\sqrt{3}}{3}\)
(5)\(\displaystyle \frac{5}{4}\pi\)
\(\displaystyle \sin\frac{5}{4}\pi=-\frac{\sqrt{2}}{2}\)
\(\displaystyle \cos\frac{5}{4}\pi=-\frac{\sqrt{2}}{2}\)
\(\displaystyle \tan\frac{5}{4}\pi=1\)
(6)\(\displaystyle \frac{4}{3}\pi\)
\(\displaystyle \sin\frac{4}{3}\pi=-\frac{\sqrt{3}}{2}\)
\(\displaystyle \cos\frac{4}{3}\pi=-\frac{1}{2}\)
\(\displaystyle \tan\frac{4}{3}\pi=\sqrt{3}\)
三角関数の相互関係
【三角関数の相互関係】
\(\sin^2\theta+\cos^2\theta=1\)
\(\displaystyle \tan\theta=\frac{\sin\theta}{\cos\theta}\)
\(\displaystyle 1+\tan^2\theta=\frac{1}{\cos^2\theta}\)
【例題】\(\theta\)が第3象限にあり、\(\displaystyle \sin\theta=-\frac{2}{3}\)のとき\(\cos\theta,\tan\theta\)の値を求めなさい。
\(\cos^2\theta=1-\sin^2\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =1-\left(-\frac{2}{3}\right)^2\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{5}{9}\)
\(\cos\theta<0\)より、
\(\displaystyle \cos\theta=-\frac{\sqrt{5}}{3}\)
\(\displaystyle \tan\theta=-\frac{2}{3}÷(-\frac{\sqrt{5}}{3})\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{2}{\sqrt{5}}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{2\sqrt{5}}{5}\)
三角関数の性質
【\(θ+2n\pi\)の三角関数】
\(\sin(θ+2n\pi)=\sinθ\)\(\cos(θ+2n\pi)=\cosθ\)
\(\tan(θ+2n\pi)=\tanθ\)
\(n\)は整数とする。
【負角の公式】
\(\sin(-θ)=-\sinθ\)\(\cos(-θ)=\cosθ\)
\(\tan(-θ)=-\tanθ\)
【補角の公式】
\(\sin(\pi-θ)=\sinθ\)\(\cos(\pi-θ)=-\cosθ\)
\(\tan(\pi-θ)=-\tanθ\)
【余角の公式】
\(\displaystyle \sin\left(\frac{\pi}{2}-θ\right)=\cosθ\)\(\displaystyle \cos\left(\frac{\pi}{2}-θ\right)=\sinθ\)
\(\displaystyle \tan\left(\frac{\pi}{2}-θ\right)=-\frac{1}{\tanθ}\)
【例題】次の数を計算しなさい。
(1)\(\displaystyle \sin\frac{5}{2}\pi\)
\(\displaystyle =\sin\frac{\pi}{2}\)
\(=1\)
(2)\(\displaystyle \cos\frac{7}{3}\pi\)
\(\displaystyle =\cos\frac{\pi}{3}\)
\(\displaystyle =\frac{1}{2}\)
(3)\(\displaystyle \tan\frac{4}{3}\pi\)
\(\displaystyle =\tan\frac{\pi}{3}\)
\(=\sqrt{3}\)
(4)\(\sinθ\cos(-θ)+\sin(\pi-θ)\cos(\pi+θ)\)
\(=\sinθ\cosθ+\sinθ・(-\cosθ)\)
\(=\sinθ\cosθ-\sinθ\cosθ\)
\(=0\)
(5)\(\displaystyle \frac{\cos(\pi-θ)\tan(-θ)\tan(\pi+θ)}{\sin(\pi+θ)\tan(\pi-θ)}\)
\(\displaystyle =\frac{-\cosθ・(-\tanθ)・\tanθ}{-\sinθ・(-\tanθ)}\)
\(\displaystyle =\frac{\cosθ\tanθ}{\sinθ}\)
\(\displaystyle =\frac{\cosθ\sinθ}{\sinθ\cosθ}\)
\(=1\)
(6)\(\displaystyle \sin\left(\frac{\pi}{2}-θ\right)\cosθ-\sinθ\cos\left(\frac{\pi}{2}+θ\right)\)
\(=\cosθ\cosθ-\sinθ・(-\sinθ)\)
\(=\cos^2θ+\sin^2θ\)
\(=1\)
(7)\(\displaystyle \frac{\tanθ\tan(\frac{\pi}{2}-θ)\sin(\frac{\pi}{2}+θ)}{\tan(\frac{\pi}{2}+θ)\cos(\frac{\pi}{2}-θ)}\)
\(\displaystyle =\frac{\tanθ・\frac{1}{\tanθ}・\cosθ}{-\frac{1}{\tanθ}\sin(θ)}\)
\(\displaystyle =-\frac{\tanθ\cosθ}{\sinθ}\)
\(\displaystyle =-\frac{\sinθ\cosθ}{\cosθ\sinθ}\)
\(=-1\)