Question

Найдите sin2a , если ​

Answer 1

Ответ: sin2α=-0,96.

Объяснение:

$\displaystyle\\cos\alpha =-\frac{4}{5} \ \ \ \ \ \ \frac{\pi }{2} < \alpha < \pi \ \ \ \ \ \ sin2\alpha =?\\\\sin^2+cos^2\alpha =1\\\\sin^2\alpha =1-cos^2\alpha =1-(-\frac{4}{5})^2=1-\frac{16}{25}=\frac{25*1-16}{25} =\frac{9}{25}.\\\\ sin\alpha =б\sqrt{\frac{9}{25} } =б\frac{3}{5}.\\\\\frac{\pi }{2} < \alpha < \pi \ \ \ \ \ \ \Rightarrow\\\\sin\alpha =\frac{3}{5} .\\\\sin2\pi =2*sin\alpha* cos\alpha =2*\frac{3}{5}*(-\frac{4}{5} )=-\frac{2*3*4}{5*5}=-\frac{24}{25}=-0,96.$