積を和に，和を積に直す（B）

(1) $\cos 75^\circ\cos 15^\circ=\dfrac{1}{2}(\cos 90^\circ+\cos 60^\circ)=\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}$ (2) $\cos A-\cos B=-2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}$ より \[ \cos 105^\circ-\cos 15^\circ=-2\sin 60^\circ\sin 45^\circ=-2\cdot\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}=-\frac{\sqrt{6}}{2} \] (3) $\sin 20^\circ\sin 40^\circ=-\dfrac{1}{2}(\cos 60^\circ-\cos 20^\circ)=\dfrac{1}{2}\left(\cos 20^\circ-\dfrac{1}{2}\right)$ であるから \[ (\text{与式})=\frac{1}{2}\cos 20^\circ\sin 80^\circ-\frac{1}{4}\sin 80^\circ \] $\cos 20^\circ\sin 80^\circ=\dfrac{1}{2}(\sin 100^\circ+\sin 60^\circ)=\dfrac{1}{2}\left(\sin 80^\circ+\dfrac{\sqrt{3}}{2}\right)$ より \[ (\text{与式})=\frac{1}{4}\sin 80^\circ+\frac{\sqrt{3}}{8}-\frac{1}{4}\sin 80^\circ=\frac{\sqrt{3}}{8} \]