Question

УМОЛЯЮ СРОЧНО ХЕЛП!!!!!!!! ЗНАЙДІТЬ sin a (НА ФОТО!!)​

Answer 1

$\displaystyle\bf\\\pi < \alpha < \frac{3\pi }{2} \ \ \ \Rightarrow \ \ \ Sin\alpha < 0\\\\\\Sin\alpha =-\sqrt{1-Cos^{2} \alpha } =-\sqrt{1-\Big(-\frac{3}{5} \Big)^{2} } =\\\\\\=-\sqrt{1-\frac{9}{25} } =-\sqrt{\frac{16}{25} }=-\frac{4}{5} \\\\\\Otvet \ : \ a)$