Question

вычислите sin a если cosa = -4/5, 180

Answer 1

Ответ: sinα=3/5.

Объяснение:

$\displaystyle\\cos\alpha =-\frac{4}{5} \ \ \ \ \ \ 90^0 < \alpha < 180^0\ \ \ \ \ \ sin\alpha =?\\\\sin^2\alpha +cos^2\alpha =1\\\\sin^2\alpha =1-cos^2\alpha \\\\sin^2\alpha =1-(-\frac{4}{5})^2=1-\frac{16}{25}=\frac{25-16}{25} =\frac{9}{25}.\\\\ sin\alpha =б\sqrt{\frac{9}{25} } =б\frac{3}{5}\\\\ 90^0 < \alpha < 180^0\ \ \ \ \ \ \Rightarrow\\\\sin\alpha =\frac{3}{5} .$