$\sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}}$, $\sin^2{\frac{\pi}{12}}$, $\cos^2{\frac{5}{12}\pi}$ の値を求める。解析学三角関数倍角の公式半角の公式三角関数の値2025/5/221. 問題の内容sinπ8cosπ8\sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}}sin8πcos8π, sin2π12\sin^2{\frac{\pi}{12}}sin212π, cos2512π\cos^2{\frac{5}{12}\pi}cos2125π の値を求める。2. 解き方の手順(1) sinπ8cosπ8\sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}}sin8πcos8πの計算2倍角の公式 sin2θ=2sinθcosθ\sin{2\theta} = 2\sin{\theta}\cos{\theta}sin2θ=2sinθcosθ を利用する。sinπ8cosπ8=12⋅2sinπ8cosπ8=12sin(2⋅π8)=12sinπ4=12⋅22=24\sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}} = \frac{1}{2} \cdot 2\sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}} = \frac{1}{2}\sin{(2 \cdot \frac{\pi}{8})} = \frac{1}{2}\sin{\frac{\pi}{4}} = \frac{1}{2}\cdot\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4}sin8πcos8π=21⋅2sin8πcos8π=21sin(2⋅8π)=21sin4π=21⋅22=42(2) sin2π12\sin^2{\frac{\pi}{12}}sin212πの計算半角の公式 sin2θ2=1−cosθ2\sin^2{\frac{\theta}{2}} = \frac{1-\cos{\theta}}{2}sin22θ=21−cosθ を利用する。sin2π12=sin2π/62=1−cosπ62=1−322=2−34\sin^2{\frac{\pi}{12}} = \sin^2{\frac{\pi/6}{2}} = \frac{1 - \cos{\frac{\pi}{6}}}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2-\sqrt{3}}{4}sin212π=sin22π/6=21−cos6π=21−23=42−3(3) cos2512π\cos^2{\frac{5}{12}\pi}cos2125πの計算半角の公式 cos2θ2=1+cosθ2\cos^2{\frac{\theta}{2}} = \frac{1+\cos{\theta}}{2}cos22θ=21+cosθ を利用する。cos2512π=cos25π/62=1+cos5π62=1−322=2−34\cos^2{\frac{5}{12}\pi} = \cos^2{\frac{5\pi/6}{2}} = \frac{1 + \cos{\frac{5\pi}{6}}}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2-\sqrt{3}}{4}cos2125π=cos225π/6=21+cos65π=21−23=42−33. 最終的な答えsinπ8cosπ8=24\sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}} = \frac{\sqrt{2}}{4}sin8πcos8π=42sin2π12=2−34\sin^2{\frac{\pi}{12}} = \frac{2-\sqrt{3}}{4}sin212π=42−3cos2512π=2−34\cos^2{\frac{5}{12}\pi} = \frac{2-\sqrt{3}}{4}cos2125π=42−3