05Jan

АлгебрSin690°ctg1200°-1/2tg750°

другвокруг Алгебра 2 19

АлгебрSin690°ctg1200°-1/2tg750°

Posted by другвокруг | Posted at Jan 05, 2014 | Categories: Алгебра

Answers

CEREDA
CEREDA

[latex]sin690^circ cdot ctg1200^circ - frac{1}{2} cdot tg750^circ =\\=sin(720^circ -30^circ)cdot ctg(4cdot 360^circ -240^circ)- frac{1}{2}cdot tg(720^circ+30^circ)=\\=sin(-30^circ)cdot ctg(-240^circ )- frac{1}{2} cdot tg30^circ=\\=-sin30^circ cdot (-ctg(360^circ-120^circ))- frac{1}{2}cdot frac{sqrt3}{3} =\\=-frac{1}{2}cdot ctg(180^circ-60^circ)- frac{sqrt3}{6}=-frac{1}{2}cdot (-ctg60^circ )- frac{sqrt3}{6}=[/latex] [latex]=frac{1}{2} cdot frac{sqrt3}{3}-frac{sqrt3}{6}=frac{sqrt3}{6}-frac{sqrt3}{6}=0[/latex]

Jan 05, 2014 06:44
Puet49
Puet49

sin690*ctg1200-1/2*tg750=sin(2*360-30)*ctg(180*7-60)-1/2tg(180*4+30)= =-sin30*(-ctg60)-1/2*tg30=-1/2*(-√3/3)-1/2*√3/3=√3/6-√3/6=0

Jan 05, 2014 06:45

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