$\cos\alpha=\dfrac{1}{2}$
Bấm $\text{SHIFT}\to \cos \to \dfrac{1}{2}$
Kết quả $\dfrac{\pi}{3}$, chưa thoả $\dfrac{5\pi}{2}<\alpha<2\pi$ nên lấy $2\pi-\dfrac{\pi}{3}=\dfrac{5\pi}{3}$ do $\begin{cases} \cos x=\cos(-x)\\ \cos(x+k2\pi)=\cos x\end{cases}$
Bấm $\sin\dfrac{5\pi}{3}$ được $\dfrac{-\sqrt3}{2}$