3. Найдите значения выражений: a) √2sin cos + cosπsin; 6) arccos(sin(arctg0)) B) tg (arccos(-)+ arcsin(-2)+4arctg1);

a) $$\sqrt{2}\sin{\frac{\pi}{4}}\cos{\frac{\pi}{3}} + \cos{\frac{\pi}{4}}\sin{\frac{\pi}{6}} = \sqrt{2} \cdot \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{2}{4} + \frac{\sqrt{2}}{4} = \frac{1}{2} + \frac{\sqrt{2}}{4} = \frac{2 + \sqrt{2}}{4}$$

б) $$\arccos(\sin(\arctan(0)))$$. Так как $$\arctan(0) = 0$$, то $$\arccos(\sin(0)) = \arccos(0) = \frac{\pi}{2}$$.

в) $$\tan(\arccos(-\frac{1}{2}) + \arcsin(-\frac{\sqrt{3}}{2}) + 4\arctan(1))$$. $$\arccos(-\frac{1}{2}) = \frac{2\pi}{3}$$, $$\arcsin(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3}$$, $$\arctan(1) = \frac{\pi}{4}$$. Тогда $$\tan(\frac{2\pi}{3} - \frac{\pi}{3} + 4 \cdot \frac{\pi}{4}) = \tan(\frac{\pi}{3} + \pi) = \tan(\frac{4\pi}{3}) = \tan(\pi + \frac{\pi}{3}) = \tan(\frac{\pi}{3}) = \sqrt{3}$$

Ответ: а) $$\frac{2 + \sqrt{2}}{4}$$, б) $$\frac{\pi}{2}$$, в) $$\sqrt{3}$$