Question

найдите tg2a если sina=12/13, пи/2<a<пи

Answer 1

$bfdisplaystyle cos(a)=pmsqrt{1-sin^{2}(a)}=pmsqrt{1-(frac{12}{13})^{2}}=pmsqrt{frac{169-144}{169}}=pmfrac{5}{13}=frac{5}{13}\\\\ tg(2a)=frac{sin(2a)}{cos(2a)}=frac{2sin(a)cos(a)}{1-2sin^{2}(a)}=frac{2cdotfrac{12}{13}cdotfrac{5}{13}}{1-2cdot(frac{12}{13})^{2}}=frac{frac{120}{169}}{frac{169-288}{169}}=frac{120cdot 169}{(-119)cdot 169}=-frac{120}{119}$