高専の数学解説

第13章 13.2

(1)

cos⁡(θ+π4)=cos⁡θ⋅cos⁡π4−sin⁡θ⋅sin⁡π4=12(cos⁡θ−sin⁡θ)\cos\left(\theta+\frac{\pi}{4}\right) = \cos\theta\cdot\cos\frac{\pi}{4} - \sin\theta\cdot\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}(\cos\theta-\sin\theta)

(2)

sin⁡(π3−θ)=sin⁡π3⋅cos⁡θ−cos⁡π3⋅sin⁡θ=12(3cos⁡θ−sin⁡θ)\sin\left(\frac{\pi}{3}-\theta\right) = \sin\frac{\pi}{3}\cdot\cos\theta - \cos\frac{\pi}{3}\cdot\sin\theta = \frac{1}{2}(\sqrt{3}\cos\theta-\sin\theta)

(3)

tan⁡(θ+π6)=tan⁡θ+tan⁡π61−tan⁡θ⋅tan⁡π6=tan⁡θ+131−tan⁡θ⋅13=3tan⁡θ+13−tan⁡θ\tan\left(\theta+\frac{\pi}{6}\right) = \frac{\tan\theta+\tan\dfrac{\pi}{6}}{1-\tan\theta\cdot\tan\dfrac{\pi}{6}} = \frac{\tan\theta+\dfrac{1}{\sqrt{3}}}{1-\tan\theta\cdot\dfrac{1}{\sqrt{3}}} = \frac{\sqrt{3}\tan\theta+1}{\sqrt{3}-\tan\theta}

(4)

cos⁡(θ−π4)=cos⁡θ⋅cos⁡π4+sin⁡θ⋅sin⁡π4=12(cos⁡θ+sin⁡θ)\cos\left(\theta-\frac{\pi}{4}\right) = \cos\theta\cdot\cos\frac{\pi}{4} + \sin\theta\cdot\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}(\cos\theta+\sin\theta)

(5)

sin⁡(θ−π6)=cos⁡θ⋅cos⁡π6−sin⁡θ⋅sin⁡π6=12(3sin⁡θ−cos⁡θ)\sin\left(\theta-\frac{\pi}{6}\right) = \cos\theta\cdot\cos\frac{\pi}{6} - \sin\theta\cdot\sin\frac{\pi}{6} = \frac{1}{2}(\sqrt{3}\sin\theta-\cos\theta)

(6)

tan⁡(θ+π4)=tan⁡θ+tan⁡π41−tan⁡θ⋅tan⁡π4=1+tan⁡θ1−tan⁡θ\tan\left(\theta+\frac{\pi}{4}\right) = \frac{\tan\theta+\tan\dfrac{\pi}{4}}{1-\tan\theta\cdot\tan\dfrac{\pi}{4}} = \frac{1+\tan\theta}{1-\tan\theta}

解説: r31bn1z