加法定理で 15° の倍数の角の三角関数の値を求める（B）

(1) $\sin 15^\circ=\sin(45^\circ-30^\circ)=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{6}-\sqrt{2}}{4}$ (2) $\cos 105^\circ=\cos(60^\circ+45^\circ)=\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}-\sqrt{6}}{4}$ (3) $\tan 165^\circ=\tan(120^\circ+45^\circ)=\dfrac{-\sqrt{3}+1}{1+\sqrt{3}}=\dfrac{(1-\sqrt{3})^2}{(1+\sqrt{3})(1-\sqrt{3})}=\dfrac{4-2\sqrt{3}}{-2}=-2+\sqrt{3}$