Answer
$\dfrac{3}{5}$
Work Step by Step
Suppose $\theta =\sin^{-1} (\dfrac{-4}{5})$
This gives: $\sin \theta=-\dfrac{4}{5}$
Since, $ r=\sqrt {x^2+y^2}$
This implies that $ x=\sqrt {r^2-y^2}=\sqrt {5^2-4^2}=3$
Therefore, we have $\cos \theta =\cos [\sin^{-1} (\dfrac{-4}{5})]=\dfrac{3}{5}$
