Вычислите: а) 3 cos 60° − 2 sin 30° + 6 ctg 60° − 2 ctg 30°; б) sin (−30°) + cos (−60°) − 2 tg (−30°) ctg (−60°); в) 5 sin (−45°) + 5 cos (−45°) − √3 tg (−30°) + sin (−30°).

а) \(\cos 60^\circ=\frac12\), \(\sin 30^\circ=\frac12\), \(\operatorname{ctg}60^\circ=\frac1{\sqrt3}\), \(\operatorname{ctg}30^\circ=\sqrt3\).

\[3\cdot\frac12-2\cdot\frac12+6\cdot\frac1{\sqrt3}-2\sqrt3=\frac12.\]

б) \(\sin(-30^\circ)=-\frac12\), \(\cos(-60^\circ)=\frac12\), \(\operatorname{tg}(-30^\circ)=-\frac1{\sqrt3}\), \(\operatorname{ctg}(-60^\circ)=-\frac1{\sqrt3}\).

\[-\frac12+\frac12-2\cdot\left(-\frac1{\sqrt3}\right)\cdot\left(-\frac1{\sqrt3}\right)=-\frac23.\]

в) \(\sin(-45^\circ)=-\frac{\sqrt2}{2}\), \(\cos(-45^\circ)=\frac{\sqrt2}{2}\), \(\operatorname{tg}(-30^\circ)=-\frac1{\sqrt3}\), \(\sin(-30^\circ)=-\frac12\).

\[5\left(-\frac{\sqrt2}{2}\right)+5\frac{\sqrt2}{2}-\sqrt3\left(-\frac1{\sqrt3}\right)-\frac12=\frac12.\]

Ответ: а) \(\frac12\); б) \(-\frac23\); в) \(\frac12\).