$\theta = -60^\circ$ のときの $\sin \theta$, $\cos \theta$, $\tan \theta$ の値を求める問題です。幾何学三角関数sincostan角度2025/7/161. 問題の内容θ=−60∘\theta = -60^\circθ=−60∘ のときの sinθ\sin \thetasinθ, cosθ\cos \thetacosθ, tanθ\tan \thetatanθ の値を求める問題です。2. 解き方の手順θ=−60∘\theta = -60^\circθ=−60∘ のとき、sinθ=sin(−60∘)\sin \theta = \sin (-60^\circ)sinθ=sin(−60∘)cosθ=cos(−60∘)\cos \theta = \cos (-60^\circ)cosθ=cos(−60∘)tanθ=tan(−60∘)\tan \theta = \tan (-60^\circ)tanθ=tan(−60∘)をそれぞれ計算します。sin(−x)=−sin(x)\sin(-x) = -\sin(x)sin(−x)=−sin(x)cos(−x)=cos(x)\cos(-x) = \cos(x)cos(−x)=cos(x)tan(−x)=−tan(x)\tan(-x) = -\tan(x)tan(−x)=−tan(x)であることと、sin(60∘)=32\sin(60^\circ) = \frac{\sqrt{3}}{2}sin(60∘)=23、cos(60∘)=12\cos(60^\circ) = \frac{1}{2}cos(60∘)=21、tan(60∘)=3\tan(60^\circ) = \sqrt{3}tan(60∘)=3 を用います。sin(−60∘)=−sin(60∘)=−32\sin (-60^\circ) = -\sin (60^\circ) = -\frac{\sqrt{3}}{2}sin(−60∘)=−sin(60∘)=−23cos(−60∘)=cos(60∘)=12\cos (-60^\circ) = \cos (60^\circ) = \frac{1}{2}cos(−60∘)=cos(60∘)=21tan(−60∘)=−tan(60∘)=−3\tan (-60^\circ) = -\tan (60^\circ) = -\sqrt{3}tan(−60∘)=−tan(60∘)=−33. 最終的な答えsinθ=−32\sin \theta = -\frac{\sqrt{3}}{2}sinθ=−23cosθ=12\cos \theta = \frac{1}{2}cosθ=21tanθ=−3\tan \theta = -\sqrt{3}tanθ=−3