Question

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

Answer 1

$\bf\displaystyle cos(a)=\pm\sqrt{1-sin^{2}(a)}=\pm\sqrt{1-(\frac{12}{13})^{2}}=\pm\sqrt{\frac{169-144}{169}}=\pm\frac{5}{13}=\frac{5}{13}\\\\\\\\ tg(2a)=\frac{sin(2a)}{cos(2a)}=\frac{2sin(a)cos(a)}{1-2sin^{2}(a)}=\frac{2\cdot\frac{12}{13}\cdot\frac{5}{13}}{1-2\cdot(\frac{12}{13})^{2}}=\frac{\frac{120}{169}}{\frac{169-288}{169}}=\frac{120\cdot 169}{(-119)\cdot 169}=-\frac{120}{119}$