【高校数学Ⅱ】4-2-2 二倍角の公式｜問題集
1.\(\displaystyle \cos\alpha=\frac{1}{3}\)のとき、次の値を求めなさい。ただし、\(\displaystyle \frac{3\pi}{2}<\alpha<2\pi\)とする。
(1)\(\sin\alpha\)
\(\displaystyle \sin^2\alpha=1-\left(\frac{1}{3}\right)^2\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ =\frac{8}{9}\)
\(\displaystyle \frac{3\pi}{2}<\alpha<2\pi\)より、
\(\displaystyle \sin\alpha=-\frac{2\sqrt{2}}{3}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ =\frac{8}{9}\)
\(\displaystyle \frac{3\pi}{2}<\alpha<2\pi\)より、
\(\displaystyle \sin\alpha=-\frac{2\sqrt{2}}{3}\)
(2)\(\sin2\alpha\)
\(=2\sin\alpha\cos\alpha\)
\(\displaystyle =2・\left(-\frac{2\sqrt{2}}{3}\right)・\frac{1}{3}\)
\(\displaystyle =-\frac{4\sqrt{2}}{9}\)
\(\displaystyle =2・\left(-\frac{2\sqrt{2}}{3}\right)・\frac{1}{3}\)
\(\displaystyle =-\frac{4\sqrt{2}}{9}\)
(3)\(\cos2\alpha\)
\(=\cos^2\alpha-\sin^2\alpha\)
\(\displaystyle =\left(\frac{1}{3}\right)^2-\left(-\frac{2\sqrt{2}}{3}\right)^2\)
\(\displaystyle =-\frac{7}{9}\)
\(\displaystyle =\left(\frac{1}{3}\right)^2-\left(-\frac{2\sqrt{2}}{3}\right)^2\)
\(\displaystyle =-\frac{7}{9}\)
(4)\(\tan2\alpha\)
\(\displaystyle =\frac{\sin2\alpha}{\cos2\alpha}\)
\(\displaystyle =\frac{-\frac{4\sqrt{2}}{9}}{-\frac{7}{9}}\)
\(\displaystyle =\frac{4\sqrt{2}}{7}\)
\(\displaystyle =\frac{-\frac{4\sqrt{2}}{9}}{-\frac{7}{9}}\)
\(\displaystyle =\frac{4\sqrt{2}}{7}\)
2.\(\displaystyle \sin\alpha=\frac{3}{5}\)のとき、次の値を求めなさい。ただし、\(\displaystyle \frac{\pi}{2}<\alpha<\pi\)とする。
(1)\(\cos\alpha\)
\(\displaystyle \cos^2\alpha=1-\left(\frac{3}{5}\right)^2\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ =\frac{16}{25}\)
\(\displaystyle \frac{\pi}{2}<\alpha<\pi\)より、
\(\displaystyle \cos\alpha=-\frac{4}{5}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ =\frac{16}{25}\)
\(\displaystyle \frac{\pi}{2}<\alpha<\pi\)より、
\(\displaystyle \cos\alpha=-\frac{4}{5}\)
(2)\(\sin2\alpha\)
\(=2\sin\alpha\cos\alpha\)
\(\displaystyle =2・\frac{3}{5}・\left(-\frac{4}{5}\right)\)
\(\displaystyle =-\frac{24}{25}\)
\(\displaystyle =2・\frac{3}{5}・\left(-\frac{4}{5}\right)\)
\(\displaystyle =-\frac{24}{25}\)
(3)\(\cos2\alpha\)
\(=\cos^2\alpha-\sin^2\alpha\)
\(\displaystyle =\left(-\frac{4}{5}\right)^2-\left(\frac{3}{5}\right)^2\)
\(\displaystyle =\frac{7}{25}\)
\(\displaystyle =\left(-\frac{4}{5}\right)^2-\left(\frac{3}{5}\right)^2\)
\(\displaystyle =\frac{7}{25}\)
(4)\(\displaystyle \sin\frac{\alpha}{2}\)
\(\displaystyle \sin^2\frac{\alpha}{2}=\frac{1-\cos\alpha}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{1+\frac{4}{5}}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{9}{10}\)
\(\displaystyle \frac{\pi}{2}<\alpha<\pi\)より、
\(\displaystyle \sin\frac{\alpha}{2}=\frac{3\sqrt{10}}{10}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{1+\frac{4}{5}}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{9}{10}\)
\(\displaystyle \frac{\pi}{2}<\alpha<\pi\)より、
\(\displaystyle \sin\frac{\alpha}{2}=\frac{3\sqrt{10}}{10}\)
3.\(\displaystyle \sin\alpha=-\frac{4}{5}\)のとき、次の値を求めなさい。ただし、\(\displaystyle \frac{3\pi}{2}<\alpha<2\pi\)とする。
(1)\(\cos\alpha\)
\(\displaystyle \cos^2\alpha=1-\left(-\frac{4}{5}\right)^2\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ =\frac{9}{25}\)
\(\displaystyle \frac{3\pi}{2}<\alpha<2\pi\)より、
\(\displaystyle \cos\alpha=\frac{3}{5}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ =\frac{9}{25}\)
\(\displaystyle \frac{3\pi}{2}<\alpha<2\pi\)より、
\(\displaystyle \cos\alpha=\frac{3}{5}\)
(2)\(\displaystyle \sin\frac{\alpha}{2}\)
\(\displaystyle \sin^2\frac{\alpha}{2}=\frac{1-\cos\alpha}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{1-\frac{3}{5}}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{5}\)
\(\displaystyle \frac{3\pi}{2}<\alpha<2\pi\)より、
\(\displaystyle \sin\frac{\alpha}{2}=\frac{\sqrt{5}}{5}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{1-\frac{3}{5}}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{5}\)
\(\displaystyle \frac{3\pi}{2}<\alpha<2\pi\)より、
\(\displaystyle \sin\frac{\alpha}{2}=\frac{\sqrt{5}}{5}\)
(3)\(\displaystyle \cos\frac{\alpha}{2}\)
\(\displaystyle \cos^2\frac{\alpha}{2}=\frac{1+\cos\alpha}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{1+\frac{3}{5}}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{4}{5}\)
\(\displaystyle \frac{3\pi}{2}<\alpha<2\pi\)より、
\(\displaystyle \cos\frac{\alpha}{2}=-\frac{2\sqrt{5}}{5}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{1+\frac{3}{5}}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{4}{5}\)
\(\displaystyle \frac{3\pi}{2}<\alpha<2\pi\)より、
\(\displaystyle \cos\frac{\alpha}{2}=-\frac{2\sqrt{5}}{5}\)
(4)\(\displaystyle \tan\frac{\alpha}{2}\)
\(\displaystyle =\frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}\)
\(\displaystyle =\frac{\frac{\sqrt{5}}{5}}{-\frac{2\sqrt{5}}{5}}\)
\(\displaystyle =-\frac{1}{2}\)
\(\displaystyle =\frac{\frac{\sqrt{5}}{5}}{-\frac{2\sqrt{5}}{5}}\)
\(\displaystyle =-\frac{1}{2}\)
4.次の値を求めなさい。
(1)\(\tan\alpha=3\)のとき、\(\tan2\alpha\)を求めなさい。
\(\tan2\alpha\)
\(\displaystyle =\frac{2\tan\alpha}{1-\tan^2\alpha}\)
\(\displaystyle =\frac{2・3}{1-3^2}\)
\(\displaystyle =-\frac{3}{4}\)
\(\displaystyle =\frac{2\tan\alpha}{1-\tan^2\alpha}\)
\(\displaystyle =\frac{2・3}{1-3^2}\)
\(\displaystyle =-\frac{3}{4}\)
(2)\(\displaystyle \cos\alpha=-\frac{2}{3}\)のとき、\(\displaystyle \tan\frac{\alpha}{2}\)を求めなさい。ただし、\(\displaystyle \frac{\pi}{2}<\alpha<\pi\)とする。
\(\displaystyle \tan^2\frac{\alpha}{2}\)
\(\displaystyle =\frac{1-\cos\alpha}{1+\cos\alpha}\)
\(\displaystyle =\frac{1+\frac{2}{3}}{1-\frac{2}{3}}\)
\(=5\)
\(\displaystyle \frac{\pi}{2}<\alpha<\pi\)より、
\(\displaystyle \tan\frac{\alpha}{2}=\sqrt{5}\)
\(\displaystyle =\frac{1-\cos\alpha}{1+\cos\alpha}\)
\(\displaystyle =\frac{1+\frac{2}{3}}{1-\frac{2}{3}}\)
\(=5\)
\(\displaystyle \frac{\pi}{2}<\alpha<\pi\)より、
\(\displaystyle \tan\frac{\alpha}{2}=\sqrt{5}\)
5.次の解を求めなさい。ただし、\(0\leqq x<2\pi\)とする。
(1)\(\cos2x+\sin x=1\)
\(1-2\sin^2x+\sin x=1\)
\(2\sin^2x-\sin x=0\)
\(\sin x(2\sin x-1)=0\)
\(\sin x=0\)より、
\(x=0,\pi\)
\(2\sin x-1=0\)より、
\(\displaystyle x=\frac{\pi}{6},\frac{5\pi}{6}\)
よって、
\(\displaystyle x=0,\frac{\pi}{6},\frac{5\pi}{6},\pi\)
\(2\sin^2x-\sin x=0\)
\(\sin x(2\sin x-1)=0\)
\(\sin x=0\)より、
\(x=0,\pi\)
\(2\sin x-1=0\)より、
\(\displaystyle x=\frac{\pi}{6},\frac{5\pi}{6}\)
よって、
\(\displaystyle x=0,\frac{\pi}{6},\frac{5\pi}{6},\pi\)
(2)\(\sin2x+\cos x=0\)
\(2\sin x\cos x+\cos x=0\)
\(\cos x(2\sin x+1)=0\)
\(\cos x=0\)より、
\(\displaystyle x=\frac{\pi}{2},\frac{3\pi}{2}\)
\(2\sin x+1=0\)より、
\(\displaystyle x=\frac{7\pi}{6},\frac{11\pi}{6}\)
よって、
\(\displaystyle x=\frac{\pi}{2},\frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}\)
\(\cos x(2\sin x+1)=0\)
\(\cos x=0\)より、
\(\displaystyle x=\frac{\pi}{2},\frac{3\pi}{2}\)
\(2\sin x+1=0\)より、
\(\displaystyle x=\frac{7\pi}{6},\frac{11\pi}{6}\)
よって、
\(\displaystyle x=\frac{\pi}{2},\frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}\)
(3)\(\sin2x=\sqrt{3}\cos x\)
\(2\sin x\cos x-\sqrt{3}\cos x=0\)
\(\cos x(2\sin x-\sqrt{3})=0\)
\(\cos x=0\)より、
\(\displaystyle x=\frac{\pi}{2},\frac{3\pi}{2}\)
\(2\sin x-\sqrt{3}=0\)より、
\(\displaystyle x=\frac{\pi}{3},\frac{2\pi}{3}\)
よって、
\(\displaystyle x=\frac{\pi}{3},\frac{\pi}{2},\frac{2\pi}{3},\frac{3\pi}{2}\)
\(\cos x(2\sin x-\sqrt{3})=0\)
\(\cos x=0\)より、
\(\displaystyle x=\frac{\pi}{2},\frac{3\pi}{2}\)
\(2\sin x-\sqrt{3}=0\)より、
\(\displaystyle x=\frac{\pi}{3},\frac{2\pi}{3}\)
よって、
\(\displaystyle x=\frac{\pi}{3},\frac{\pi}{2},\frac{2\pi}{3},\frac{3\pi}{2}\)
(4)\(\sin2x=-\sqrt{2}\cos x\)
\(2\sin x\cos x+\sqrt{2}\cos x=0\)
\(\cos x(2\sin x+\sqrt{2})=0\)
\(\cos x=0\)より、
\(\displaystyle x=\frac{\pi}{2},\frac{3\pi}{2}\)
\(2\sin x+\sqrt{2}=0\)より、
\(\displaystyle x=\frac{5\pi}{4},\frac{7\pi}{4}\)
よって、
\(\displaystyle x=\frac{\pi}{2},\frac{5\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}\)
\(\cos x(2\sin x+\sqrt{2})=0\)
\(\cos x=0\)より、
\(\displaystyle x=\frac{\pi}{2},\frac{3\pi}{2}\)
\(2\sin x+\sqrt{2}=0\)より、
\(\displaystyle x=\frac{5\pi}{4},\frac{7\pi}{4}\)
よって、
\(\displaystyle x=\frac{\pi}{2},\frac{5\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}\)
(5)\(\cos2x<3\sin x-1\)
\(1-2\sin^2x<3\sin x-1\)
\(2\sin^2x+3\sin x-2>0\)
\((2\sin x-1)(\sin x+2)>0\)
\(\sin x+2>0\)は常に成り立つので、
\(2\sin x-1>0\)
よって、
\(\displaystyle \frac{\pi}{6}< x< \frac{5\pi}{6}\)
\(2\sin^2x+3\sin x-2>0\)
\((2\sin x-1)(\sin x+2)>0\)
\(\sin x+2>0\)は常に成り立つので、
\(2\sin x-1>0\)
よって、
\(\displaystyle \frac{\pi}{6}< x< \frac{5\pi}{6}\)
(6)\(\cos2x<3\cos x+1\)
\(2\cos^2x-1<3\cos x+1\)
\(2\cos^2x-3\cos x-2<0\)
\((2\cos x+1)(\cos x-2)<0\)
\(\cos x-2<0\)は常に成り立つので、
\(2\cos x+1>0\)
よって、
\(\displaystyle 0\leqq x<\frac{2\pi}{3},\frac{4\pi}{3}< x<2\pi\)
\(2\cos^2x-3\cos x-2<0\)
\((2\cos x+1)(\cos x-2)<0\)
\(\cos x-2<0\)は常に成り立つので、
\(2\cos x+1>0\)
よって、
\(\displaystyle 0\leqq x<\frac{2\pi}{3},\frac{4\pi}{3}< x<2\pi\)
次の学習に進もう！