Question

1) tg2п/3•sin3п/2+3ctgп/3-2cosп/4 2)2sin45cos45+3cos60•tg180 3)3tg3п/4•cos3п/4-sin3п/4+2 4)3sin60cos60+2cos90•tg45-3

Answer 1

$1)\quad tg \frac{2\pi}{3} \cdot sin \frac{3\pi }{2} +3ctg \frac{\pi}{3} -2cos\frac{\pi}{4}= \\\\=tg(\pi -\frac{\pi}{3})\cdot sin(\pi +\frac{\pi}{2})+3\cdot \frac{\sqrt3}{3}-2\cdot \frac{\sqrt2}{2}=\\\\=-tg\frac{\pi}{3}\cdot (-sin\frac{\pi}{2})+\sqrt3-\sqrt2=-\sqrt3\cdot (-1)+\sqrt3-\sqrt2=\\\\=2\sqrt3-\sqrt2=\sqrt2(\sqrt6-1)\\\\2)\quad 2sin45\cdot cos45+3cos60\cdot tg180=sin90+3\cdot \frac{1}{2}\cdot 0=1$

$3)\quad 3tg\frac{3\pi}{4}\cdot cos\frac{3\pi}{4}-sin\frac{3\pi}{4}+2=\\\\=3tg(\pi-\frac{\pi}{4})\cdot cos(\pi-\frac{\pi}{4})-sin(\pi-\frac{\pi}{4})+2=\\\\=3\cdot (-tg\frac{\pi}{4})\cdot (-cos\frac{\pi}{4})-sin\frac{\pi}{4}+2=\\\\=-3\cdot 1 \cdot (-\frac{\sqrt2}{2})-\frac{\sqrt2}{2}+2=2\cdot \frac{\sqrt2}{2}+2=\sqrt2+2=\sqrt2(1+\sqrt2)\\\\4)\quad 3sin60\cdot cos60+2cos90\cdot tg45-3=\\\\=\frac{3}{2}\cdot sin120+2\cdot 0\cdot \frac{\sqrt2}{2}-3=\\\\=\frac{3}{2}\cdot sin(180-60)+0-3=\frac{3}{2}\cdot sin60-3=$

$=\frac{3}{2}\cdot \frac{\sqrt3}{2}-3=3\sqrt3-3=3(\sqrt3-1)$