4-2-2 二倍角の公式(要点)

【二倍角の公式】

\(\sin2\alpha=2\sin\alpha\cos\alpha\)
\(\cos2\alpha=\cos^2\alpha-\sin^2\alpha\)
\(\ \ \ \ \ \ \ \ \ \ \ =1-2\sin^2\alpha\)
\(\ \ \ \ \ \ \ \ \ \ \ =2\cos^2\alpha-1\)
\(\displaystyle \tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}\)

【例題】\(\displaystyle \sin\alpha=\frac{3}{5}\)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、
\(\displaystyle \cos\alpha=-\frac{4}{5}\)

(2)\(\sin2\alpha\)

\(=2\sin\alpha\cos\alpha\)
\(\displaystyle =2・\frac{3}{5}・\left(-\frac{4}{5}\right)\)
\(\displaystyle =-\frac{24}{25}\)

(3)\(\cos2\alpha\)

\(=\cos^2\alpha-\sin^2\alpha\)
\(\displaystyle =\left(-\frac{4}{5}\right)^2-\left(\frac{3}{5}\right)^2\)
\(\displaystyle =\frac{7}{25}\)

(4)\(\tan2\alpha\)

\(\displaystyle =\frac{2\tan\alpha}{1-\tan^2\alpha}\)
\(\displaystyle =\frac{2・(-\frac{3}{4})}{1-(-\frac{3}{4})^2}\)
\(\displaystyle =-\frac{24}{7}\)

【半角の公式】

\(\displaystyle \sin^2\frac{\alpha}{2}=\frac{1-\cos\alpha}{2}\)
\(\displaystyle \cos^2\frac{\alpha}{2}=\frac{1+\cos\alpha}{2}\)
\(\displaystyle \tan^2\frac{\alpha}{2}=\frac{1-\cos\alpha}{1+\cos\alpha}\)

【例題】次の値を求めなさい。ただし、\(0<\alpha<\pi\)とする。

(1)\(\displaystyle \sin\frac{3\pi}{8}\)

\(\displaystyle \sin^2\frac{3\pi}{8}=\frac{1-\cos\frac{3\pi}{4}}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1-(-\frac{\sqrt{2}}{2})}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{2+\sqrt{2}}{4}\)
\(0<\alpha<\pi\)より、
\(\displaystyle \sin\frac{3\pi}{8}=\frac{\sqrt{2+\sqrt{2}}}{2}\)

(2)\(\displaystyle \cos\frac{\pi}{8}\)

\(\displaystyle \cos^2\frac{\pi}{8}=\frac{1+\cos\frac{\pi}{4}}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1+\frac{\sqrt{2}}{2}}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{2+\sqrt{2}}{4}\)
\(0<\alpha<\pi\)より、
\(\displaystyle \cos\frac{\pi}{8}=\frac{\sqrt{2+\sqrt{2}}}{2}\)

(3)\(\displaystyle \tan\frac{\pi}{8}\)

\(\displaystyle \tan^2\frac{\pi}{8}=\frac{1-\cos\frac{\pi}{4}}{1+\cos\frac{\pi}{4}}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{2-\sqrt{2}}{2+\sqrt{2}}\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ =3-2\sqrt{2}\)
\(0<\alpha<\pi\)より、
\(\displaystyle \tan\frac{\pi}{8}=\sqrt{3-2\sqrt{2}}\)
\(\ \ \ \ \ \ \ \ \ \ \ =\sqrt{2}-\sqrt{1}\)
\(\ \ \ \ \ \ \ \ \ \ \ =\sqrt{2}-1\)

【例題】次の解を求めなさい。ただし、\(0\leqq x<2\pi\)とする。

(1)\(\sin2x-\cos x=0\)

\(2\sin x\cos x-\cos x=0\)
\(\cos x(2\sin x-1)=0\)
\(\cos x=0\)より、
\(\displaystyle x=\frac{\pi}{2},\frac{3\pi}{2}\)
\(2\sin x-1=0\)より、
\(\displaystyle x=\frac{\pi}{6},\frac{5\pi}{6}\)
よって、
\(\displaystyle x=\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}\)

(2)\(\cos2x-3\cos x+2\leqq0\)

\(2\cos^2x-1-3\cos x+2\leqq0\)
\(2\cos^2x-3\cos x+1\leqq0\)
\((2\cos x-1)(\cos x-1)\leqq0\)
\(\displaystyle \frac{1}{2}\leqq\cos x\leqq1\)
よって、
\(\displaystyle 0\leqq x\leqq\frac{\pi}{3},\frac{5\pi}{3}\leqq x\leqq2\pi\)