Задача 31. Найдите \(\sin \,\alpha \), если \({\text{ctg}}\,\alpha = -\dfrac{4}{3}\) и \(\alpha \in \left( {\dfrac{{3\pi }}{2};\,2\pi } \right)\)
Решение
Воспользуемся формулой: \(1 + {\rm{ct}}{{\rm{g}}^2}\alpha = \dfrac{1}{{{{\sin }^2}\alpha }}.\)
\(1 + {\left( {\dfrac{4}{3}} \right)^2} = \dfrac{1}{{{{\sin }^2}\alpha }}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\alpha = \dfrac{9}{{25}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \alpha = \dfrac{3}{5},\,\,\,}\\{\sin \alpha = -\dfrac{3}{5}.}\end{array}} \right.\)
Так как \(\alpha \, \in \,\left( {\dfrac{{3\pi }}{2};2\pi } \right)\) (IV четверть), то \(\sin \alpha < 0\), то есть \(\sin \alpha = -\dfrac{3}{5} = -0,6.\)
Ответ: \(-0,6.\)