Question

Алгебра найти cos a, если sin а =-4÷5; П<а<3П÷2

Answer 1

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