## Giúp mi câu 90 và 91 ạ

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Giúp mi câu 90 và 91 ạ

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1 year 2021-01-22T01:38:55+00:00 2 Answers 8 views 0

$$\begin{array}{l} 90,\\ 90^\circ < \alpha < 180^\circ \Rightarrow \left\{ \begin{array}{l} \cos \alpha < 0\\ \sin \alpha > 0 \end{array} \right.\\ {\sin ^2}\alpha + {\cos ^2}\alpha = 1\\ \sin \alpha > 0 \Rightarrow \sin \alpha = \sqrt {1 – {{\cos }^2}\alpha } = \frac{1}{5}\\ \tan 2\alpha = \frac{{\sin 2\alpha }}{{\cos 2\alpha }} = \frac{{2\sin \alpha .\cos \alpha }}{{2{{\cos }^2}\alpha – 1}} = \frac{{ – 4\sqrt 6 }}{{23}}\\ \cot \left( { – a} \right) = \frac{{\cos \left( { – a} \right)}}{{\sin \left( { – a} \right)}} = \frac{{\cos a}}{{ – \sin a}} = 2\sqrt 6 \\ 91,\\ \frac{{3\pi }}{2} < \alpha < 2\pi \Rightarrow \left\{ \begin{array}{l} \sin \alpha < 0\\ \cos \alpha > 0 \end{array} \right.\\ {\sin ^2}\alpha + {\cos ^2}\alpha = 1\\ \cos \alpha > 0 \Rightarrow \cos \alpha = \sqrt {1 – {{\sin }^2}\alpha } = \frac{4}{5}\\ \tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{ – 3}}{4}\\ \cot 2\alpha = \frac{{\cos 2\alpha }}{{\sin 2\alpha }} = \frac{{2{{\cos }^2}\alpha – 1}}{{2\sin \alpha .\cos \alpha }} = \frac{{ – 7}}{{24}}\\ \sin \left( {\pi – \alpha } \right) = \sin \alpha = – \frac{3}{5}\\ \cos \left( {\frac{\pi }{2} – \alpha } \right) = \sin \alpha = – \frac{3}{5} \end{array}$$