Cho sin α = 4/5 và pi/2 < α < pi . Tính cos α

Ta có:

\[\begin{array}{l}{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1 \Leftrightarrow {\left( {\frac{4}{5}} \right)^2} + {\cos ^2}\alpha  = 1 \Leftrightarrow {\cos ^2}\alpha  = 1 - {\left( {\frac{4}{5}} \right)^2} \Leftrightarrow {\cos ^2}\alpha  = \frac{9}{{25}}\\ \Rightarrow \left[ \begin{array}{l}\cos \alpha  = \frac{3}{5}\\\cos \alpha  = \frac{{ - 3}}{5}\end{array} \right.\end{array}\]

Vì \[\frac{\pi }{2} < \alpha  < \pi \] nên \[\cos \alpha  < 0 \Rightarrow \cos \alpha  = \frac{{ - 3}}{5}\]