$\theta = \frac{\pi}{6}$ のとき、$\sin\theta$, $\cos\theta$, $\tan\theta$ の値をそれぞれ求める問題です。幾何学三角関数sincostan角度2025/7/281. 問題の内容θ=π6\theta = \frac{\pi}{6}θ=6π のとき、sinθ\sin\thetasinθ, cosθ\cos\thetacosθ, tanθ\tan\thetatanθ の値をそれぞれ求める問題です。2. 解き方の手順θ=π6\theta = \frac{\pi}{6}θ=6π は 30∘30^\circ30∘ に相当します。三角関数の定義より、sin(π6)=12\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}sin(6π)=21cos(π6)=32\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}cos(6π)=23tan(π6)=sin(π6)cos(π6)=1232=13=33\tan\left(\frac{\pi}{6}\right) = \frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}tan(6π)=cos(6π)sin(6π)=2321=31=333. 最終的な答えsinθ=12\sin\theta = \frac{1}{2}sinθ=21cosθ=32\cos\theta = \frac{\sqrt{3}}{2}cosθ=23tanθ=33\tan\theta = \frac{\sqrt{3}}{3}tanθ=33