$\cos(\alpha + 270^\circ - y)$ を展開せよ。幾何学三角関数加法定理三角関数の変換2025/7/311. 問題の内容cos(α+270∘−y)\cos(\alpha + 270^\circ - y)cos(α+270∘−y) を展開せよ。2. 解き方の手順まず、270∘−y270^\circ - y270∘−y を AAA と置くと、与えられた式は cos(α+A)\cos(\alpha + A)cos(α+A) となります。cos(α+A)\cos(\alpha + A)cos(α+A) の加法定理は次のようになります。cos(α+A)=cosαcosA−sinαsinA\cos(\alpha + A) = \cos \alpha \cos A - \sin \alpha \sin Acos(α+A)=cosαcosA−sinαsinAここで、AAA を 270∘−y270^\circ - y270∘−y に戻すと、cos(α+270∘−y)=cosαcos(270∘−y)−sinαsin(270∘−y)\cos (\alpha + 270^\circ - y) = \cos \alpha \cos(270^\circ - y) - \sin \alpha \sin(270^\circ - y)cos(α+270∘−y)=cosαcos(270∘−y)−sinαsin(270∘−y)cos(270∘−y)=cos(270∘)cos(y)+sin(270∘)sin(y)=0⋅cos(y)+(−1)⋅sin(y)=−siny\cos(270^\circ - y) = \cos(270^\circ)\cos(y) + \sin(270^\circ)\sin(y) = 0 \cdot \cos(y) + (-1) \cdot \sin(y) = -\sin ycos(270∘−y)=cos(270∘)cos(y)+sin(270∘)sin(y)=0⋅cos(y)+(−1)⋅sin(y)=−sinysin(270∘−y)=sin(270∘)cos(y)−cos(270∘)sin(y)=(−1)⋅cos(y)−0⋅sin(y)=−cosy\sin(270^\circ - y) = \sin(270^\circ)\cos(y) - \cos(270^\circ)\sin(y) = (-1) \cdot \cos(y) - 0 \cdot \sin(y) = -\cos ysin(270∘−y)=sin(270∘)cos(y)−cos(270∘)sin(y)=(−1)⋅cos(y)−0⋅sin(y)=−cosyこれらの結果を代入すると、cos(α+270∘−y)=cosα(−siny)−sinα(−cosy)=−cosαsiny+sinαcosy=sinαcosy−cosαsiny\cos(\alpha + 270^\circ - y) = \cos \alpha (-\sin y) - \sin \alpha (-\cos y) = -\cos \alpha \sin y + \sin \alpha \cos y = \sin \alpha \cos y - \cos \alpha \sin ycos(α+270∘−y)=cosα(−siny)−sinα(−cosy)=−cosαsiny+sinαcosy=sinαcosy−cosαsinyさらに、sinαcosy−cosαsiny \sin \alpha \cos y - \cos \alpha \sin ysinαcosy−cosαsiny は sin(α−y)\sin(\alpha - y)sin(α−y) となります。3. 最終的な答えsin(α−y)\sin(\alpha - y)sin(α−y)