【高校数学Ⅱ】4-2-1 加法定理｜問題集

\(=\sin(135°+30°)\)
\(=\sin135°\cos30°+\cos135°\sin30°\)
\(\displaystyle =\frac{1}{\sqrt{2}}・\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}・\frac{1}{2}\)
\(\displaystyle =\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\(\displaystyle =\frac{\sqrt{6}-\sqrt{2}}{4}\)

\(=\cos(45°+30°)\)
\(=\cos45°\cos30°-\sin45°\sin30°\)
\(\displaystyle =\frac{1}{\sqrt{2}}・\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}・\frac{1}{2}\)
\(\displaystyle =\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\(\displaystyle =\frac{\sqrt{6}-\sqrt{2}}{4}\)

\(\displaystyle =\tan\left(\frac{\pi}{4}-\frac{\pi}{6}\right)\)
\(\displaystyle =\frac{\tan\frac{\pi}{4}-\tan\frac{\pi}{6}}{1+\tan\frac{\pi}{4}\tan\frac{\pi}{6}}\)
\(\displaystyle =\frac{1-\frac{1}{\sqrt{3}}}{1+1・\frac{1}{\sqrt{3}}}\)
\(\displaystyle =\frac{\sqrt{3}-1}{\sqrt{3}+1}\)
\(=2-\sqrt{3}\)

\(=\sin(45°-30°)\)
\(=\sin45°\cos30°-\cos45°\sin30°\)
\(\displaystyle =\frac{1}{\sqrt{2}}・\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}・\frac{1}{2}\)
\(\displaystyle =\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\(\displaystyle =\frac{\sqrt{6}-\sqrt{2}}{4}\)

\(=\cos(150°+45°)\)
\(=\cos150°\cos45°-\sin150°\sin45°\)
\(\displaystyle =-\frac{\sqrt{3}}{2}・\frac{1}{\sqrt{2}}-\frac{1}{2}・\frac{1}{\sqrt{2}}\)
\(\displaystyle =\frac{-\sqrt{3}-1}{2\sqrt{2}}\)
\(\displaystyle =-\frac{\sqrt{6}+\sqrt{2}}{4}\)

\(\displaystyle \sin\alpha=\frac{2}{3}\)より、\(\displaystyle \cos^2\alpha=\frac{5}{9}\)
\(\alpha\)は第\(2\)象限より、
\(\displaystyle \cos\alpha=-\frac{\sqrt{5}}{3}\)
\(\displaystyle \cos\beta=\frac{3}{5}\)より、\(\displaystyle \sin^2\beta=\frac{16}{25}\)
\(\beta\)は第\(1\)象限より、
\(\displaystyle \sin\beta=\frac{4}{5}\)
よって、
\(\sin(\alpha-\beta)\)
\(=\sin\alpha\cos\beta-\cos\alpha\sin\beta)\)
\(\displaystyle =\frac{2}{3}・\frac{3}{5}+\frac{\sqrt{5}}{3}・\frac{4}{5}\)
\(\displaystyle =\frac{6}{15}+\frac{4\sqrt{5}}{15}\)
\(\displaystyle =\frac{6+4\sqrt{5}}{15}\)

\(\cos(\alpha+\beta)\)
\(=\cos\alpha\cos\beta-\sin\alpha\sin\beta)\)
\(\displaystyle =-\frac{\sqrt{5}}{3}・\frac{3}{5}-\frac{2}{3}・\frac{4}{5}\)
\(\displaystyle =-\frac{3\sqrt{5}}{15}-\frac{8}{15}\)
\(\displaystyle =-\frac{8+3\sqrt{5}}{15}\)

\(\displaystyle \sin\alpha=\frac{2}{3}\)より、\(\displaystyle \cos^2\alpha=\frac{5}{9}\)
\(\alpha\)は第\(1\)象限より、
\(\displaystyle \cos\alpha=\frac{\sqrt{5}}{3}\)
\(\displaystyle \sin\beta=\frac{4}{5}\)より、\(\displaystyle \cos^2\beta=\frac{9}{25}\)
\(\beta\)は第\(2\)象限より、
\(\displaystyle \cos\beta=-\frac{3}{5}\)
よって、
\(\sin(\alpha+\beta)\)
\(=\sin\alpha\cos\beta+\cos\alpha\sin\beta)\)
\(\displaystyle =\frac{2}{3}\left(-\frac{3}{5}\right)+\frac{\sqrt{5}}{3}・\frac{4}{5}\)
\(\displaystyle =-\frac{6}{15}+\frac{4\sqrt{5}}{15}\)
\(\displaystyle =\frac{-6+4\sqrt{5}}{15}\)

\(\cos(\alpha+\beta)\)
\(=\cos\alpha\cos\beta-\sin\alpha\sin\beta)\)
\(\displaystyle =\frac{\sqrt{5}}{3}・\left(-\frac{3}{5}\right)-\frac{2}{3}・\frac{4}{5}\)
\(\displaystyle =-\frac{3\sqrt{5}}{15}-\frac{8}{15}\)
\(\displaystyle =-\frac{8+3\sqrt{5}}{15}\)

\(y=2x-1\)の傾き\(2\)なので、\(\tan\alpha=2\)とおく。
\(\displaystyle y=\frac{1}{3}x+1\)の傾き\(\displaystyle \frac{1}{3}\)なので、\(\displaystyle \tan\beta=\frac{1}{3}\)とおく。
よって、
\(\tan\theta\)
\(=\tan(\alpha-\beta)\)
\(\displaystyle =\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\)
\(\displaystyle =\frac{2-\frac{1}{3}}{1+2・\frac{1}{3}}\)
\(=1\)
\(\displaystyle 0\leqq\theta\leqq\frac{\pi}{2}\)より、
\(\displaystyle \theta=\frac{\pi}{4}\)

\(y=2x\)の傾き\(2\)なので、\(\tan\alpha=2\)とおく。
\(x-3y+1=0\)の傾き\(\displaystyle \frac{1}{3}\)なので、\(\displaystyle \tan\beta=\frac{1}{3}\)とおく。
よって、
\(\tan\theta\)
\(=\tan(\alpha-\beta)\)
\(\displaystyle =\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\)
\(\displaystyle =\frac{2-\frac{1}{3}}{1+2・\frac{1}{3}}\)
\(=1\)
\(\displaystyle 0\leqq\theta\leqq\frac{\pi}{2}\)より、
\(\displaystyle \theta=\frac{\pi}{4}\)