Ответ:
\(\cos(-3\pi-\beta)=-\cos\beta\), \(\sin(\beta-\frac\pi2)=-\cos\beta\), \(\cos(\beta+\pi)=-\cos\beta\).
\[\frac{2(-\cos\beta)-\cos\beta}{3(-\cos\beta)}=\frac{-3\cos\beta}{-3\cos\beta}=1.\]
Ответ: 1.
\(\cos(-3\pi-\beta)=-\cos\beta\), \(\sin(\beta-\frac\pi2)=-\cos\beta\), \(\cos(\beta+\pi)=-\cos\beta\).
\[\frac{2(-\cos\beta)-\cos\beta}{3(-\cos\beta)}=\frac{-3\cos\beta}{-3\cos\beta}=1.\]
Ответ: 1.