## Precalculus (6th Edition) Blitzer

Identity for $\sin \left( C-D \right)$ is expressed as $\sin C\cos D-\cos C\sin D$.
From the difference formula of sines, the sine of the difference between two angles, say A and B, is expressed as, $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ Thus, for angles C and D, $\sin \left( C-D \right)=\sin C\cos D-\cos C\sin D$