Question

Найдите tg2x,если sinx=5/6
​2π ​

Answer 1

Ответ:

Объяснение:

$\displaystyle sin~x = \frac{5}{6} \\\\ cos~ x = \pm \sqrt{1-\bigg(\frac{5}{6}\bigg)^2 }=\pm \frac{\sqrt{11} }{6} \\\\\\ tg ~ x=\frac{sin ~x}{cos~ x} =\frac{\dfrac{5}{6} }{\pm \cfrac{\sqrt{11} }{6}} = \pm \frac{5}{\sqrt{11} }$

Воспользуемся формулой


$tg ~ 2x =\dfrac{2tg~x}{1-tg^2~ x}$

При
$tg ~ x = \pm \dfrac{\sqrt{11} }{6}$  

$tg ~2x= \dfrac{2\cdot \pm \cfrac{5}{\sqrt{6} } }{1-\bigg(\pm \cfrac{5}{\sqrt{6} } \bigg)^2} =\cfrac{\pm \cfrac{10}{ \sqrt{6} } }{\cfrac{19}{6} } = \pm \dfrac{95}{3\sqrt{6} }$