角を変換して三角関数の値を求める（B）

(1) (ア) $\cos\dfrac{17}{4}\pi=\cos\left(\dfrac{\pi}{4}+4\pi\right)=\dfrac{\sqrt{2}}{2}$ (イ) $\sin\left(-\dfrac{7}{3}\pi\right)=-\sin\dfrac{7}{3}\pi=-\sin\dfrac{\pi}{3}=-\dfrac{\sqrt{3}}{2}$ (ウ) $\tan\left(-\dfrac{5}{6}\pi\right)=\tan\left(-\dfrac{5}{6}\pi+\pi\right)=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}$ (2) $\cos(\pi+\theta)=-\cos\theta$，$\sin\left(\dfrac{\pi}{2}+\theta\right)=\cos\theta$，$\cos(\pi-\theta)=-\cos\theta$，$\sin\left(\dfrac{3}{2}\pi+\theta\right)=-\cos\theta$ であるから \[ (\text{与式})=-\cos\theta+\cos\theta-\cos\theta-\cos\theta=-2\cos\theta \]