Question

x∈(0,π/2) si sinx=1/2$\sqrt{2+ \sqrt{3$ calculati tg 2x

Answer 1

$sin^2x+cos^2x=1\Rightarrow cos^2x=1-\dfrac{2+\sqrt3}{4}=\dfrac{2-\sqrt3}{4}\Rightarrow$

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

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

$tgx=2+\sqrt3$

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

$=\dfrac{1}{\sqrt3}$