Question

dados, sen x=1/4,cos x=1/5 e tg x=3, no intervalo 180°<x<270°,calcule : sen 2x cos 2x tg 2x.

Answer 1

O intervalo dado pertence ao terceiro quadrante, portanto senx e cosx são negativos.

$senx=-1/4 \ \ \ \ cosx=-1/5\\ sen2x=2*-senx*-cosx\\ sen2x=2*-1/4*-1/5\\ sen2x=2/20\\ sen2x=1/10$

$cos2x=-cos^2x-(-sen^2x)\\ cos2x=(-1/5)^2-(-1/4)^2\\ cos2x=1/25-1/16\\ mmc(25,16)=400\\ cos2x=(16-25)/400\\ cos2x=-9/400$

$tg2x= \frac{2tgx}{1-tg^2x}\\\\ tgx= \frac{senx}{cosx} \\\\ tgx= \frac{-1/4} {-1/5}\\\\ tgx= -1/4*-5=5/4\\ tg^2x=(5/4)^2=25/16\\\\ tg2x= \frac{2*5/4}{1-25/16}\\\\ tg2x= \frac{5/2}{-9/16}\\\\ tg2x=5/2*16/-9\\ tg2x=80/-18\\ tg2x=-40/9$

Resposta: sen2x=1/10   cos2x=-9/400   tg2x=-40/9