Question

Вычислить:
$\sqrt{13}*sin \alpha$, если tg $\alpha$ = -1,5 и ∈ (2π;3π)

Answer 1

$tg \alpha =-1,5\; ,\; \; 2\pi \ \textless \ \alpha \ \textless \ 3\pi \\\\tg \alpha \ \textless \ 0\; \; \Rightarrow \; \; \alpha \in (\frac{5\pi }{2},3\pi )\; \; \Rightarrow \; \; sin \alpha\ \textgreater \ 0\\\\1+ctg^2 \alpha = \frac{1}{sin^2 \alpha } \; \; \to \; \; 1+ \frac{1}{tg^2 \alpha } = \frac{1}{sin^2 \alpha }\; ,\\\\\frac{tg^2 \alpha +1}{tg^2 \alpha }=\frac{1}{sin^2 \alpha }\; \; \Rightarrow \; \; \; sin^2 \alpha =\frac{tg^2 \alpha }{1+tg^2 \alpha }= \frac{2,25}{1+2,25}=\frac{2,25}{3,25}=\frac{9}{13}\; \Rightarrow$

$sin \alpha =+\sqrt{\frac{9}{13}}= \frac{3}{\sqrt{13}} \\\\\sqrt{13}\cdot sin \alpha =\sqrt{13}\cdot \frac{3}{\sqrt{13}} =3$