第13章 13.7

(1)

$\sin\frac{\pi}{4}\cos\frac{\pi}{6} = \frac{1}{2}\left\{\sin\left(\frac{\pi}{4}+\frac{\pi}{6}\right) + \sin\left(\frac{\pi}{4}-\frac{\pi}{6}\right)\right\} = \frac{1}{2}\left(\sin\frac{5}{12}\pi + \sin\frac{\pi}{12}\right)$

(2)

$\cos\frac{\pi}{6}\cos\frac{\pi}{3} = \frac{1}{2}\left\{\cos\left(\frac{\pi}{6}+\frac{\pi}{3}\right) + \cos\left(\frac{\pi}{6}-\frac{\pi}{3}\right)\right\} = \frac{1}{2}\left(\cos\frac{\pi}{2} + \cos\frac{\pi}{6}\right) = \frac{1}{2}\cos\frac{\pi}{6}$

(3)

$\sin\frac{2}{3}\pi\sin\frac{\pi}{3} = -\frac{1}{2}\left\{\cos\left(\frac{2}{3}\pi+\frac{\pi}{3}\right) - \cos\left(\frac{2}{3}\pi-\frac{\pi}{3}\right)\right\} = -\frac{1}{2}\left(\cos\pi - \cos\frac{\pi}{3}\right) = \frac{1}{2}\left(1 + \cos\frac{\pi}{3}\right)$

(4)

$\cos 7x \cdot \sin 5x = \frac{1}{2}\left(\sin 12x - \sin 2x\right)$

(5)

$\cos 2x \cdot \cos 4x = \frac{1}{2}\left(\cos 6x + \cos 2x\right)$

(6)

$\sin 4x \cdot \sin 2x = -\frac{1}{2}\left(\cos 6x - \cos 2x\right)$