Ответ:
\(\sin(60^\circ-\beta)=\sin60^\circ\cos\beta-\cos60^\circ\sin\beta\).
Подставим значения: \(\sin60^\circ=\frac{\sqrt{3}}{2}\), \(\cos60^\circ=\frac{1}{2}\).
\(\sin(60^\circ-\beta)=\frac{\sqrt{3}}{2}\cos\beta-\frac{1}{2}\sin\beta\).
\(\cos(\beta-30^\circ)=\cos\beta\cos30^\circ+\sin\beta\sin30^\circ\).
Так как \(\cos30^\circ=\frac{\sqrt{3}}{2}\), \(\sin30^\circ=\frac{1}{2}\), то
\(\cos(\beta-30^\circ)=\frac{\sqrt{3}}{2}\cos\beta+\frac{1}{2}\sin\beta\).
Ответ: а) \(\frac{\sqrt{3}}{2}\cos\beta-\frac{1}{2}\sin\beta\); б) \(\frac{\sqrt{3}}{2}\cos\beta+\frac{1}{2}\sin\beta\).
