Question

Значение выражения соs a sin a tg a если cos a = корень -3/4

Answer 1

$cos \alpha =- \frac{3}{4}\ \textless \ 0\; \; \to \; \; \alpha \in (\frac{\pi}{2}, \frac{3\pi}{2})\\\\1)\; \; Esli\; \alpha \in ( \frac{\pi}{2} ,\pi)\; ,\; to\; \; sin \alpha \ \textgreater \ 0,\; tg \alpha \ \textless \ 0:\\\\sin^2 \alpha =1-cos^2 \alpha =1-\frac{9}{16}= \frac{7}{9}\; ,\; \; sin \alpha =+\frac{\sqrt7}{3} \\\\tg \alpha = \frac{sin \alpha }{cos \alpha } =\frac{\sqrt7/3}{-3/4} =- \frac{4\sqrt7}{9} \\\\2)\; \; Esli\; \alpha \in (\pi ,\frac{3\pi}{2})\; ,\; to\; sin \alpha \ \textless \ 0\; ,\; tg \alpha \ \textgreater \ 0:$

$sin \alpha =- \frac{\sqrt7}{3} \; ,\; \; tg \alpha =\frac{4\sqrt7}{3}$