Упр.16.6 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) sin(-30°)-2tg(-45°)+cos(-45°);
2) 5tg 0+2sin(-?/6)-3ctg(-?/4)+4cos(-?/2);
3) tg(-?/3)ctg(-?/6)+2cos(-?)+4sin^2(-?/3).

1. $$\sin(-30^\circ)-2\tg(-45^\circ)+\cos(-45^\circ)$$
   $$=-\sin 30^\circ+2\tg 45^\circ+\cos 45^\circ$$
   $$=-\frac12+2\cdot 1+\frac{\sqrt2}{2}$$
   $$=-\frac12+\frac42+\frac{\sqrt2}{2}=\frac{3+\sqrt2}{2}.$$

2. $$5\tg 0+2\sin\left(-\frac{\pi}{6}\right)-3\ctg\left(-\frac{\pi}{4}\right)+4\cos\left(-\frac{\pi}{2}\right)$$
   $$=5\cdot 0-2\sin\frac{\pi}{6}+3\ctg\frac{\pi}{4}+4\cos\frac{\pi}{2}$$
   $$=0-2\cdot\frac12+3\cdot 1+4\cdot 0=-1+3=2.$$

3. $$\tg\left(-\frac{\pi}{3}\right)\ctg\left(-\frac{\pi}{6}\right)+2\cos(-\pi)+4\sin^2\left(-\frac{\pi}{3}\right)$$
   $$=\tg\left(-\frac{\pi}{3}\right)\ctg\left(-\frac{\pi}{6}\right)+2\cos\pi+4\sin^2\left(\frac{\pi}{3}\right)$$
   $$=(-\sqrt3)\cdot(-\sqrt3)+2\cdot(-1)+4\cdot\left(\frac{\sqrt3}{2}\right)^2$$
   $$=3-2+4\cdot\frac34=1+3=4.$$

Ответ

1) $$\frac{3+\sqrt2}{2}$$; 2) $$2$$; 3) $$4$$.