Răspunsuri

2014-03-21T08:59:30+02:00
sin^2x+cos^2x=1\Rightarrow cos^2x=1-\dfrac{2+\sqrt3}{4}=\dfrac{2-\sqrt3}{4}\Rightarrow

\Rightarrow cosx=\dfrac{\sqrt{2-\sqrt3}}{2} Am luat semnul + deoarece x este in primul cadran.

tgx=\dfrac{sinx}{cosx}=\sqrt{\dfrac{2+\sqrt3}{2-\sqrt3}} Dupa rationalizarea numitorului

tgx=2+\sqrt3

tg2x=\dfrac{2tgx}{1-tg^2x}=\dfrac{2(2+\sqrt3)}{1-(2+\sqrt3)^2}=\dfrac{2(2+\sqrt3)}{-6-4\sqrt3}=\dfrac{2(2+\sqrt3)}{-2\sqrt3(\sqrt3+2)}=

=\dfrac{1}{\sqrt3}