Question

Помогите сделать очень надо, пожалуйста!

Answer 1

$tg\beta =-2\\\\\beta \in 2\; koord.\; ygol\; \; \to \; \; sin\beta >0\; ,\; \; cos\beta <0\; \; ,\; \; ctg\beta <0\\\\\\1+tg^2\beta =\frac{1}{cos^2\beta }\; \; \to \; \; cos^2\beta =\frac{1}{1+tg^2\beta }=\frac{1}{1+4}=\frac{1}{5}\; ,\; \; cos\beta =-\frac{1}{\sqrt5}=-\frac{\sqrt5}{5}\\\\sin\beta =tg\beta \cdot cos\beta =-2\cdot (-\frac{1}{\sqrt5})=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}\\\\ctg\beta =\frac{1}{tg\beta }=-\frac{1}{2}$