【高校数学Ⅱ】4-1-2 三角関数の性質｜問題集
1.次の値を求めなさい。
(1)\(\displaystyle \cos\frac{5}{4}\pi\)
\(\displaystyle =-\frac{\sqrt{2}}{2}\)
(2)\(\displaystyle \sin\frac{11}{6}\pi\)
\(\displaystyle =-\frac{1}{2}\)
(3)\(\displaystyle \tan\frac{4}{3}\pi\)
\(\displaystyle =\sqrt{3}\)
(4)\(\displaystyle \sin\left(-\frac{\pi}{6}\right)\)
\(\displaystyle =-\frac{1}{2}\)
(5)\(\displaystyle \cos\left(-\frac{13}{6}\pi\right)\)
\(\displaystyle =\cos\left(-\frac{\pi}{6}\right)\)
\(\displaystyle =\frac{\sqrt{3}}{2}\)
\(\displaystyle =\frac{\sqrt{3}}{2}\)
(6)\(\displaystyle \tan\left(-\frac{9}{4}\pi\right)\)
\(\displaystyle =\tan\left(-\frac{\pi}{4}\right)\)
\(\displaystyle =-1\)
\(\displaystyle =-1\)
2.次の\(\theta\)について、\(\sin\theta,\cos\theta,\tan\theta\)の値を求めなさい。
(1)\(\displaystyle \frac{5}{3}\pi\)
\(\displaystyle \sin\frac{5}{3}\pi=-\frac{\sqrt{3}}{2}\)
\(\displaystyle \cos\frac{5}{3}\pi=\frac{1}{2}\)
\(\displaystyle \tan\frac{5}{3}\pi=-\sqrt{3}\)
\(\displaystyle \cos\frac{5}{3}\pi=\frac{1}{2}\)
\(\displaystyle \tan\frac{5}{3}\pi=-\sqrt{3}\)
(2)\(\displaystyle -\frac{2}{3}\pi\)
\(\displaystyle \sin\left(-\frac{2}{3}\pi\right)=-\frac{\sqrt{3}}{2}\)
\(\displaystyle \cos\left(-\frac{2}{3}\pi\right)=-\frac{1}{2}\)
\(\displaystyle \tan\left(-\frac{2}{3}\pi\right)=\sqrt{3}\)
\(\displaystyle \cos\left(-\frac{2}{3}\pi\right)=-\frac{1}{2}\)
\(\displaystyle \tan\left(-\frac{2}{3}\pi\right)=\sqrt{3}\)
(3)\(\displaystyle -\frac{3}{4}\pi\)
\(\displaystyle \sin\left(-\frac{3}{4}\pi\right)=-\frac{\sqrt{2}}{2}\)
\(\displaystyle \cos\left(-\frac{3}{4}\pi\right)=-\frac{\sqrt{2}}{2}\)
\(\displaystyle \tan\left(-\frac{3}{4}\pi\right)=1\)
\(\displaystyle \cos\left(-\frac{3}{4}\pi\right)=-\frac{\sqrt{2}}{2}\)
\(\displaystyle \tan\left(-\frac{3}{4}\pi\right)=1\)
3.次の値を求めなさい。
(1)\(\theta\)が第\(4\)象限にあり、\(\displaystyle \sin\theta=-\frac{1}{3}\)のとき\(\cos\theta,\tan\theta\)の値を求めなさい。
\(\cos^2\theta=1-\sin^2\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =1-\left(-\frac{1}{3}\right)^2\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{8}{9}\)
\(\cos\theta>0\)より、
\(\displaystyle \cos\theta=\frac{2\sqrt{2}}{3}\)
\(\displaystyle \tan\theta=-\frac{1}{3}÷\frac{2\sqrt{2}}{3}\)
\(\displaystyle \ \ \ \ \ \ \ \ =-\frac{1}{2\sqrt{2}}\)
\(\displaystyle \ \ \ \ \ \ \ \ =-\frac{\sqrt{2}}{4}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =1-\left(-\frac{1}{3}\right)^2\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{8}{9}\)
\(\cos\theta>0\)より、
\(\displaystyle \cos\theta=\frac{2\sqrt{2}}{3}\)
\(\displaystyle \tan\theta=-\frac{1}{3}÷\frac{2\sqrt{2}}{3}\)
\(\displaystyle \ \ \ \ \ \ \ \ =-\frac{1}{2\sqrt{2}}\)
\(\displaystyle \ \ \ \ \ \ \ \ =-\frac{\sqrt{2}}{4}\)
(2)\(\theta\)が第\(3\)象限にあり、\(\tan\theta=3\)のとき\(\sin\theta,\cos\theta\)の値を求めなさい。
\(\displaystyle \cos^2\theta=\frac{1}{1+\tan^2\theta}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{1}{1+3^2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{1}{10}\)
\(\cos\theta<0\)より、
\(\displaystyle \cos\theta=-\frac{\sqrt{10}}{10}\)
\(\displaystyle \sin^2\theta=1-\cos^2\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =1-\left(-\frac{1}{\sqrt{10}}\right)^2\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{9}{10}\)
\(\sin\theta<0\)より、
\(\displaystyle \sin\theta=-\frac{3\sqrt{10}}{10}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{1}{1+3^2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{1}{10}\)
\(\cos\theta<0\)より、
\(\displaystyle \cos\theta=-\frac{\sqrt{10}}{10}\)
\(\displaystyle \sin^2\theta=1-\cos^2\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =1-\left(-\frac{1}{\sqrt{10}}\right)^2\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{9}{10}\)
\(\sin\theta<0\)より、
\(\displaystyle \sin\theta=-\frac{3\sqrt{10}}{10}\)
(3)\(\theta\)が第\(4\)象限にあり、\(\displaystyle \cos\theta=\frac{1}{3}\)のとき\(\sin\theta,\tan\theta\)の値を求めなさい。
\(\sin^2\theta=1-\cos^2\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =1-\left(\frac{1}{3}\right)^2\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{8}{9}\)
\(\sin\theta<0\)より、
\(\displaystyle \sin\theta=-\frac{2\sqrt{2}}{3}\)
\(\displaystyle \tan\theta=-\frac{2\sqrt{2}}{3}÷\frac{1}{3}\)
\(\displaystyle \ \ \ \ \ \ \ \ =-2\sqrt{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =1-\left(\frac{1}{3}\right)^2\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{8}{9}\)
\(\sin\theta<0\)より、
\(\displaystyle \sin\theta=-\frac{2\sqrt{2}}{3}\)
\(\displaystyle \tan\theta=-\frac{2\sqrt{2}}{3}÷\frac{1}{3}\)
\(\displaystyle \ \ \ \ \ \ \ \ =-2\sqrt{2}\)
4.\(\displaystyle \sin\theta+\cos\theta=\frac{1}{2}\)のとき、次の値を求めなさい。
(1)\(\sin\theta\cos\theta\)
\(\displaystyle \sin\theta+\cos\theta=\frac{1}{2}\)の両辺を\(2\)乗すると、
\(\displaystyle (\sin\theta+\cos\theta)^2=\frac{1}{4}\)
\(\displaystyle \sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta=\frac{1}{4}\)
\(\displaystyle 2\sin\theta\cos\theta=\frac{1}{4}-1\)
\(\displaystyle \sin\theta\cos\theta=-\frac{3}{8}\)
\(\displaystyle (\sin\theta+\cos\theta)^2=\frac{1}{4}\)
\(\displaystyle \sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta=\frac{1}{4}\)
\(\displaystyle 2\sin\theta\cos\theta=\frac{1}{4}-1\)
\(\displaystyle \sin\theta\cos\theta=-\frac{3}{8}\)
(2)\(\sin^3\theta+\cos^3\theta\)
\(=(\sin\theta+\cos\theta)^3-3\sin\theta\cos\theta(\sin\theta+\cos\theta)\)
\(\displaystyle =\left(\frac{1}{2}\right)^3-3・\left(-\frac{3}{8}\right)・\frac{1}{2}\)
\(\displaystyle =\frac{1}{8}+\frac{9}{16}\)
\(\displaystyle =\frac{11}{16}\)
\(\displaystyle =\left(\frac{1}{2}\right)^3-3・\left(-\frac{3}{8}\right)・\frac{1}{2}\)
\(\displaystyle =\frac{1}{8}+\frac{9}{16}\)
\(\displaystyle =\frac{11}{16}\)
(3)\(\sin\theta-\cos\theta\)
\((\sin\theta-\cos\theta)^2\)
\(=\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta\)
\(\displaystyle =1-2・\left(-\frac{3}{8}\right)\)
\(\displaystyle =\frac{7}{4}\)
\(\displaystyle \sin\theta-\cos\theta=\pm\frac{\sqrt{7}}{2}\)
\(=\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta\)
\(\displaystyle =1-2・\left(-\frac{3}{8}\right)\)
\(\displaystyle =\frac{7}{4}\)
\(\displaystyle \sin\theta-\cos\theta=\pm\frac{\sqrt{7}}{2}\)
5.等式\(\tan^2\theta-\sin^2\theta=\tan^2\theta\sin^2\theta\)を証明しなさい。
(左辺)\(=\tan^2\theta-\sin^2\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ =\sin^2\theta\left(\frac{1}{\cos^2\theta}-1\right)\)
\(\ \ \ \ \ \ \ \ =\sin^2\theta\tan^2\theta\)
よって、
\(\tan^2\theta-\sin^2\theta=\tan^2\theta\sin^2\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ =\sin^2\theta\left(\frac{1}{\cos^2\theta}-1\right)\)
\(\ \ \ \ \ \ \ \ =\sin^2\theta\tan^2\theta\)
よって、
\(\tan^2\theta-\sin^2\theta=\tan^2\theta\sin^2\theta\)
6.次の値を求めなさい。
(1)\(\displaystyle \sin\frac{15}{4}\pi\)
\(\displaystyle =\sin\frac{7}{4}\pi\)
\(\displaystyle =-\frac{\sqrt{2}}{2}\)
\(\displaystyle =-\frac{\sqrt{2}}{2}\)
(2)\(\displaystyle \tan\left(-\frac{5}{3}\pi\right)\)
\(\displaystyle =-\tan\frac{5}{3}\pi\)
\(=\sqrt{3}\)
\(=\sqrt{3}\)
(3)\(\displaystyle \sin\frac{6}{7}\pi+\cos\frac{9}{14}\pi\)
\(\displaystyle =\sin\left(\pi-\frac{\pi}{7}\right)+\cos\left(\frac{\pi}{7}+\frac{\pi}{2}\right)\)
\(\displaystyle =\sin\frac{\pi}{7}-\sin\frac{\pi}{7}\)
\(=0\)
\(\displaystyle =\sin\frac{\pi}{7}-\sin\frac{\pi}{7}\)
\(=0\)
(4)\(\displaystyle \sin\frac{100}{3}\pi\)
\(\displaystyle =\sin\frac{4}{3}\pi\)
\(\displaystyle =-\frac{\sqrt{3}}{2}\)
\(\displaystyle =-\frac{\sqrt{3}}{2}\)
(5)\(\displaystyle \tan\left(-\frac{3}{4}\pi\right)\)
\(\displaystyle =-\tan\frac{3}{4}\pi\)
\(=1\)
\(=1\)
(6)\(\displaystyle \sin\frac{3}{10}\pi+\cos\frac{4}{5}\pi\)
\(\displaystyle =\sin\left(\frac{\pi}{2}-\frac{\pi}{5}\right)+\cos\left(\pi-\frac{\pi}{5}\right)\)
\(\displaystyle =\cos\frac{\pi}{5}-\cos\frac{\pi}{5}\)
\(=0\)
\(\displaystyle =\cos\frac{\pi}{5}-\cos\frac{\pi}{5}\)
\(=0\)
次の学習に進もう！