## Precalculus (6th Edition) Blitzer

The result of $\sin \left( \alpha -\beta \right)\cos \beta +\cos \left( \alpha -\beta \right)\sin \beta$ is $\sin \alpha$.
Let us consider the given expression: $\sin \left( \alpha -\beta \right)\cos \beta +\cos \left( \alpha -\beta \right)\sin \beta$ By using the trigonometric identities, $\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta$ $\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$ Now, the above expression can be further simplified as, \begin{align} & \sin \left( \alpha -\beta \right)\cos \beta +\cos \left( \alpha -\beta \right)\sin \beta =\left\{ \left( \sin \alpha \cos \beta -\cos \alpha \sin \beta \right)\cos \beta +\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)\sin \beta \right\} \\ & =\left\{ \sin \alpha {{\cos }^{2}}\beta -\cos \alpha \sin \beta \cos \beta +\cos \alpha \cos \beta \sin \beta +\sin \alpha {{\sin }^{2}}\beta \right\} \\ & =\sin \alpha {{\cos }^{2}}\beta +\sin \alpha {{\sin }^{2}}\beta \\ & =sin\alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right) \end{align} By using the identity, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ \begin{align} & sin\alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right)=\sin \alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right) \\ & =\sin \alpha \left( 1 \right) \\ & =\sin \alpha \end{align} Thus, $\sin \left( \alpha -\beta \right)\cos \beta +\cos \left( \alpha -\beta \right)\sin \beta$ can be simplified as $\sin \alpha$. Hence, the result of $\sin \left( \alpha -\beta \right)\cos \beta +\cos \left( \alpha -\beta \right)\sin \beta$ is $\sin \alpha$.