Question

помогите

$\sin\alpha \cos \alpha = - \frac{1}{ \sqrt{5} } \alpha e(\pi2\pi)$
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Answer 1

Ответ:

$1)\ \ \ cosa=-\dfrac{1}{\sqrt5}<0\ \ ,\ \ \ \ a\in (\, \pi \, ;\, 2\pi \, )\\\\\\\boxed{sin^2a+cos^2a=1}\ \ \Rightarrow \ \ \ sin^2a=1-cos^2a=1-\dfrac{1}{25}=\dfrac{24}{25}\\\\\\a\in (\, \pi \, ;\, 2\pi \, )\ \ \ \Rightarrow \ \ \ sina<0\ \ ,\ \ sina=-\dfrac{\sqrt{24}}{5}=-\dfrac{2\sqrt6}{5}$

$2)\ \ sina=-\dfrac{\sqrt3}{2}<0\ \ ,\ \ \ \ a\in \Big(\, \pi \, ;\, \dfrac{3\pi }{2}\, \Big)\\\\\\\boxed{sin^2a+cos^2a=1}\ \ \Rightarrow \ \ \ cos^2a=1-sin^2a=1-\dfrac{3}{4}=\dfrac{1}{4}\\\\\\a\in \Big(\, \pi \, ;\, \dfrac{3\pi }{2}\, \Big)\ \ \ \Rightarrow \ \ \ cosa<0\ \ ,\ \ cosa=-\sqrt{\dfrac{1}{4}}=-\dfrac{1}{2}$