Vì \({\cos ^2}\alpha + {\sin ^2}\alpha = 1\) nên \({\cos ^2}\alpha = 1 - {\sin ^2}\alpha = 1 - {\left( { - \frac{4}{5}} \right)^2} = \frac{9}{{25}}\)
Do \(\pi < \alpha < \frac{{3\pi }}{2}\) nên \(\cos \alpha < 0\). Suy ra \(\cos \alpha = - \frac{3}{5}\)