Diketahui tan A= -½ dengan 90° < A < 180° . Nilai sin A . cos A = ...

[tex] \sin = \frac{1}{ \sqrt{5} } \\ \cos = \frac{ - 2}{ \sqrt{5} } \\ \sin( \alpha ) \times \cos( \alpha ) = \frac{1}{ \sqrt{5} } \times \frac{ - 2}{ \sqrt{5} } \\ = \frac{ - 2}{ 5 } [/tex]