530. Найти значение выражения: 1) cos 630° − sin 1470° − ctg 1125°; 2) tg 1800° − sin 495° + cos 945°; 3) 3 cos 3660° + sin (−1560°) + cos (−450°); 4) cos 4455° − cos (−945°) + tg 1035° − ctg (−1500°).

1) Используем периодичность тригонометрических функций:

\(\cos 630^\circ=\cos 270^\circ=0\), \(\sin 1470^\circ=\sin 30^\circ=\frac{1}{2}\), \(\operatorname{ctg}1125^\circ=\operatorname{ctg}45^\circ=1\).

Следовательно, \(0-\frac{1}{2}-1=-\frac{3}{2}\).

2) \(\operatorname{tg}1800^\circ=0\), \(\sin495^\circ=\sin135^\circ=\frac{\sqrt2}{2}\), \(\cos945^\circ=\cos225^\circ=-\frac{\sqrt2}{2}\).

\(0-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=-\sqrt2\).

3) \(\cos3660^\circ=\cos60^\circ=\frac12\), \(\sin(-1560^\circ)=\sin240^\circ=-\frac{\sqrt3}{2}\), \(\cos(-450^\circ)=\cos450^\circ=0\).

\(3\cdot\frac12-\frac{\sqrt3}{2}+0=\frac{3-\sqrt3}{2}\).

4) \(\cos4455^\circ=\cos135^\circ=-\frac{\sqrt2}{2}\), \(\cos(-945^\circ)=\cos945^\circ=-\frac{\sqrt2}{2}\), \(\operatorname{tg}1035^\circ=\operatorname{tg}135^\circ=-1\), \(\operatorname{ctg}(-1500^\circ)=\operatorname{ctg}120^\circ=-\frac{\sqrt3}{3}\).

\(-\frac{\sqrt2}{2}-(-\frac{\sqrt2}{2})-1-(-\frac{\sqrt3}{3})=\frac{\sqrt3}{3}-1\).

Ответ: 1) \(-\frac32\); 2) \(-\sqrt2\); 3) \(\frac{3-\sqrt3}{2}\); 4) \(\frac{\sqrt3}{3}-1\).