Answer
$\frac{{dy}}{{d\theta }} = - 1$
Work Step by Step
$$\eqalign{
& y = {\sin ^{ - 1}}\left( {\cos \theta } \right),{\text{ 0}} < \theta < \pi \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{{\sin }^{ - 1}}\left( {\cos \theta } \right)} \right] \cr
& {\text{Use }}\frac{d}{{d\theta }}\left[ {{{\sin }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{d\theta }},{\text{ let }}u = \cos \theta {\text{ }}\left( {{\text{see page 223}}} \right) \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\sqrt {1 - {{\left( {\cos \theta } \right)}^2}} }}\frac{d}{{d\theta }}\left[ {\cos \theta } \right] \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\sqrt {1 - {{\cos }^2}\theta } }}\left( { - \sin \theta } \right) \cr
& {\text{Multiply and simplify}} \cr
& \frac{{dy}}{{d\theta }} = - \frac{{\sin \theta }}{{\sqrt {1 - {{\cos }^2}\theta } }} \cr
& {\text{Use the identity }}{\cos ^2}\theta + {\sin ^2}\theta = 1 \cr
& \frac{{dy}}{{d\theta }} = - \frac{{\sin \theta }}{{\sqrt {{{\sin }^2}\theta } }} \cr
& \frac{{dy}}{{d\theta }} = - \frac{{\sin \theta }}{{\left| {\sin \theta } \right|}},{\text{ 0}} < \theta < \pi {\text{ }} \to \sin \theta > 0 \cr
& \frac{{dy}}{{d\theta }} = - \frac{{\sin \theta }}{{\sin \theta }} \cr
& \frac{{dy}}{{d\theta }} = - 1 \cr} $$
