Correct. Draw $\displaystyle{{190}}^{\circ}\usepackage{cmbright}\Huge\$ in standard position. The terminal side of $\displaystyle{{190}}^{\circ}\usepackage{cmbright}\Huge\$ lies in the third quadrant where y is negative, so $\displaystyle{\sin{{\left({{190}}^{\circ}\right)}}}={y}\usepackage{cmbright}\Huge\$ is negative.