角を変換して三角関数の値を求める（A）

(1) (ア) $\sin\dfrac{13}{6}\pi=\sin\left(\dfrac{\pi}{6}+2\pi\right)=\sin\dfrac{\pi}{6}=\dfrac{1}{2}$ (イ) $\cos\left(-\dfrac{11}{3}\pi\right)=\cos\dfrac{11}{3}\pi=\cos\left(\dfrac{5}{3}\pi+2\pi\right)=\cos\dfrac{5}{3}\pi=\dfrac{1}{2}$ (ウ) $\tan\dfrac{5}{4}\pi=\tan\left(\dfrac{\pi}{4}+\pi\right)=\tan\dfrac{\pi}{4}=1$ (2) $\sin(\pi-\theta)=\sin\theta$，$\cos\left(\dfrac{\pi}{2}+\theta\right)=-\sin\theta$，$\sin(\pi+\theta)=-\sin\theta$，$\cos\left(\dfrac{\pi}{2}-\theta\right)=\sin\theta$ であるから \[ (\text{与式})=\sin\theta-\sin\theta-\sin\theta+\sin\theta=0 \]