高専の数学解説

第13章 13.1

sin⁡α=45\sin\alpha = \dfrac{4}{5}から cos⁡α=35\cos\alpha = \dfrac{3}{5} cos⁡β=513\cos\beta = \dfrac{5}{13}から sin⁡β=1213\sin\beta = \dfrac{12}{13}

(1)

sin⁡(α+β)=sin⁡α⋅cos⁡β+cos⁡α⋅sin⁡β=45⋅513+35⋅1213=413+3665=5665\sin(\alpha+\beta) = \sin\alpha \cdot \cos\beta + \cos\alpha \cdot \sin\beta = \dfrac{4}{5}\cdot\dfrac{5}{13} + \dfrac{3}{5}\cdot\dfrac{12}{13} = \dfrac{4}{13} + \dfrac{36}{65} = \dfrac{56}{65}

(2)

cos⁡(α+β)=cos⁡α⋅cos⁡β−sin⁡α⋅sin⁡β=35⋅513−45⋅1213=1565−4865=−3365\cos(\alpha+\beta) = \cos\alpha \cdot \cos\beta - \sin\alpha \cdot \sin\beta = \dfrac{3}{5}\cdot\dfrac{5}{13} - \dfrac{4}{5}\cdot\dfrac{12}{13} = \dfrac{15}{65} - \dfrac{48}{65} = -\dfrac{33}{65}

(3)

sin⁡(β−α)=sin⁡β⋅cos⁡α−cos⁡β⋅sin⁡α=1213⋅35−513⋅45=3665−2065=1665\sin(\beta-\alpha) = \sin\beta \cdot \cos\alpha - \cos\beta \cdot \sin\alpha = \dfrac{12}{13}\cdot\dfrac{3}{5} - \dfrac{5}{13}\cdot\dfrac{4}{5} = \dfrac{36}{65} - \dfrac{20}{65} = \dfrac{16}{65}

(4)

cos⁡(β−α)=cos⁡β⋅cos⁡α+sin⁡β⋅sin⁡α=513⋅35+1213⋅45=1565+4865=6365\cos(\beta-\alpha) = \cos\beta \cdot \cos\alpha + \sin\beta \cdot \sin\alpha = \dfrac{5}{13}\cdot\dfrac{3}{5} + \dfrac{12}{13}\cdot\dfrac{4}{5} = \dfrac{15}{65} + \dfrac{48}{65} = \dfrac{63}{65}

解説: r31bn1z