4. Вычислить: a) arccos1 – arccos (−½) + arccos √3/2 б) arccos (sin π/6) в) tg (arccos(−√2/2)) г) arcsin1 – arcsin ½ + arcsin (−√3/2) д) arcsin (cos π/3) e) ctg (arcsin √3/2 + arccos½) ж) arctg √3 – arctg1 + arcctg(−√3)

a) $$\arccos 1 - \arccos \left(-\frac{1}{2}\right) + \arccos \frac{\sqrt{3}}{2} = 0 - \frac{2\pi}{3} + \frac{\pi}{6} = -\frac{4\pi}{6} + \frac{\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}$$

б) $$\arccos \left(\sin \frac{\pi}{6}\right) = \arccos \frac{1}{2} = \frac{\pi}{3}$$

в) $$\operatorname{tg} \left(\arccos \left(-\frac{\sqrt{2}}{2}\right)\right) = \operatorname{tg} \frac{3\pi}{4} = -1$$

г) $$\arcsin 1 - \arcsin \frac{1}{2} + \arcsin \left(-\frac{\sqrt{3}}{2}\right) = \frac{\pi}{2} - \frac{\pi}{6} - \frac{\pi}{3} = \frac{3\pi}{6} - \frac{\pi}{6} - \frac{2\pi}{6} = 0$$

д) $$\arcsin \left(\cos \frac{\pi}{3}\right) = \arcsin \frac{1}{2} = \frac{\pi}{6}$$

е) $$\operatorname{ctg} \left(\arcsin \frac{\sqrt{3}}{2} + \arccos \frac{1}{2}\right) = \operatorname{ctg} \left(\frac{\pi}{3} + \frac{\pi}{3}\right) = \operatorname{ctg} \frac{2\pi}{3} = -\frac{\sqrt{3}}{3}$$

ж) $$\operatorname{arctg} \sqrt{3} - \operatorname{arctg} 1 + \operatorname{arcctg} \left(-\sqrt{3}\right) = \frac{\pi}{3} - \frac{\pi}{4} + \frac{5\pi}{6} = \frac{4\pi}{12} - \frac{3\pi}{12} + \frac{10\pi}{12} = \frac{11\pi}{12}$$

Ответ:

a) $$\arccos 1 - \arccos \left(-\frac{1}{2}\right) + \arccos \frac{\sqrt{3}}{2} = -\frac{\pi}{2}$$

б) $$\arccos \left(\sin \frac{\pi}{6}\right) = \frac{\pi}{3}$$

в) $$\operatorname{tg} \left(\arccos \left(-\frac{\sqrt{2}}{2}\right)\right) = -1$$

г) $$\arcsin 1 - \arcsin \frac{1}{2} + \arcsin \left(-\frac{\sqrt{3}}{2}\right) = 0$$

д) $$\arcsin \left(\cos \frac{\pi}{3}\right) = \frac{\pi}{6}$$

е) $$\operatorname{ctg} \left(\arcsin \frac{\sqrt{3}}{2} + \arccos \frac{1}{2}\right) = -\frac{\sqrt{3}}{3}$$

ж) $$\operatorname{arctg} \sqrt{3} - \operatorname{arctg} 1 + \operatorname{arcctg} \left(-\sqrt{3}\right) = \frac{11\pi}{12}$$