【高校数学Ⅱ】4-2-1 加法定理｜要点まとめ

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

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

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

\(\cos^2\alpha\)
\(=1-\sin^2\alpha\)
\(\displaystyle =1-\left(-\frac{3}{5}\right)^2\)
\(\displaystyle =\frac{16}{25}\)
\(\alpha\)は第\(3\)象限より、
\(\displaystyle \cos\alpha=-\frac{4}{5}\)
\(\sin^2\beta\)
\(=1-\cos^2\beta\)
\(\displaystyle =1-\left(\frac{1}{3}\right)^2\)
\(\displaystyle =\frac{8}{9}\)
\(\beta\)は第\(4\)象限より、
\(\displaystyle \sin\beta=-\frac{2\sqrt{2}}{3}\)
よって、
\(\sin(\alpha+\beta)\)
\(=\sin\alpha\cos\beta+\cos\alpha\sin\beta)\)
\(\displaystyle =-\frac{3}{5}・\frac{1}{3}-\frac{4}{5}・\left(-\frac{2\sqrt{2}}{3}\right)\)
\(\displaystyle =-\frac{3}{15}+\frac{8\sqrt{2}}{15}\)
\(\displaystyle =\frac{8\sqrt{2}-3}{15}\)

\(\cos(\alpha-\beta)\)
\(=\cos\alpha\cos\beta+\sin\alpha\sin\beta)\)
\(\displaystyle =-\frac{4}{5}・\frac{1}{3}-\frac{3}{5}・\left(-\frac{2\sqrt{2}}{3}\right)\)
\(\displaystyle =-\frac{4}{15}+\frac{6\sqrt{2}}{15}\)
\(\displaystyle =\frac{6\sqrt{2}-4}{15}\)

\(=\tan(150°-45°)\)
\(\displaystyle =\frac{\tan150°-\tan45°}{1+\tan150°\tan45°}\)
\(\displaystyle =\frac{-\frac{1}{\sqrt{3}}-1}{1-\frac{1}{\sqrt{3}}・1}\)
\(\displaystyle =\frac{1+\sqrt{3}}{1-\sqrt{3}}\)
\(=-2-\sqrt{3}\)

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

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