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

$1)quad tg frac{2pi}{3} cdot sin frac{3pi }{2} +3ctg frac{pi}{3} -2cosfrac{pi}{4}= \\=tg(pi -frac{pi}{3})cdot sin(pi +frac{pi}{2})+3cdot frac{sqrt3}{3}-2cdot frac{sqrt2}{2}=\\=-tgfrac{pi}{3}cdot (-sinfrac{pi}{2})+sqrt3-sqrt2=-sqrt3cdot (-1)+sqrt3-sqrt2=\\=2sqrt3-sqrt2=sqrt2(sqrt6-1)\\2)quad 2sin45cdot cos45+3cos60cdot tg180=sin90+3cdot frac{1}{2}cdot 0=1$

$3)quad 3tgfrac{3pi}{4}cdot cosfrac{3pi}{4}-sinfrac{3pi}{4}+2=\\=3tg(pi-frac{pi}{4})cdot cos(pi-frac{pi}{4})-sin(pi-frac{pi}{4})+2=\\=3cdot (-tgfrac{pi}{4})cdot (-cosfrac{pi}{4})-sinfrac{pi}{4}+2=\\=-3cdot 1 cdot (-frac{sqrt2}{2})-frac{sqrt2}{2}+2=2cdot frac{sqrt2}{2}+2=sqrt2+2=sqrt2(1+sqrt2)\\4)quad 3sin60cdot cos60+2cos90cdot tg45-3=\\=frac{3}{2}cdot sin120+2cdot 0cdot frac{sqrt2}{2}-3=\\=frac{3}{2}cdot sin(180-60)+0-3=frac{3}{2}cdot sin60-3=$

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