Question

Найти значение тригонометрической функции tga=5/12 π

Answer 1

$tg \frac{5\pi }{12}=tg75^\circ =tg(30^\circ+45^\circ)= \frac{tg30^\circ +tg45^\circ}{1-tg30^\circ\cdot tg45^\circ}= \frac{\frac{\sqrt3}{3}+1}{1-\frac{\sqrt3}{3}}=\\\\= \frac{\sqrt3+3}{3-\sqrt3}=\frac{\sqrt3(1+\sqrt3)}{\sqrt3(\sqrt3-1)} = \frac{(1+\sqrt3)^2}{(\sqrt3-1)(\sqrt3+1)}=\frac{1+2\sqrt3+3}{3-1}=\\\\=\frac{4+2\sqrt3}{2}=2+\sqrt3$