第12章 12.5

(1)

$\sin\theta\cos\theta = \frac{\sqrt{3}}{4} \qquad \sin^2\theta\cos^2\theta = \frac{3}{16} \qquad (1-\cos^2\theta)\cos^2\theta = \frac{3}{16} \qquad \cos^4\theta - \cos^2\theta + \frac{3}{16} = 0$

$\left(\cos^2\theta - \frac{1}{4}\right)\left(\cos^2\theta - \frac{3}{4}\right) = 0 \qquad \cos^2\theta = \frac{1}{4}, \frac{3}{4} \qquad \cos\theta = \pm\frac{1}{2}, \pm\frac{\sqrt{3}}{2} \quad \text{このとき} \quad \sin\theta = \pm\frac{\sqrt{3}}{2}, \pm\frac{1}{2}$

これから

$\sin\theta + \cos\theta = \pm\frac{1+\sqrt{3}}{2}$

(2)

$\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta) = \left(\pm\frac{1+\sqrt{3}}{2}\right)\left(1 - \frac{\sqrt{3}}{4}\right) = \pm\frac{1+3\sqrt{3}}{8}$