Вычислить: [tex] \sqrt{13}*sin \alpha[/tex], если tg [tex] \alpha [/tex] = -1,5 и ∈ (2π;3π). Created by 666FATAL. algebra-ru

tex] = -1,5 и ∈ (2π;3π)...

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

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

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

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