## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 46

#### Answer

See the full explanation below.

#### Work Step by Step

Let us consider the left side of the given expression: $\sin \left( \alpha +\beta \right)\sin \left( \alpha -\beta \right)$ By using the identities of trigonometry, $\sin \,\left( \alpha +\beta \right)=\sin \,\alpha \,\cos \beta +\cos \,\alpha \,\sin \,\beta$ $\sin \,\left( \alpha -\beta \right)=\sin \,\alpha \,\cos \beta -\cos \,\alpha \,\sin \,\beta$ , the above expression can be further simplified as: \begin{align} & \sin \,\left( \alpha +\beta \right)\sin \,\left( \alpha -\beta \right)=\left( \sin \,\alpha \,\cos \beta +\cos \,\alpha \,\sin \,\beta \right)\times \left( \sin \,\alpha \,\cos \beta -\cos \,\alpha \,\sin \,\beta \right) \\ & ={{\sin }^{2}}\,\alpha \,{{\cos }^{2}}\,\beta -\sin \,\alpha \,\sin \,\beta \cos \,\alpha \,\cos \beta \\ & +\sin \,\alpha \,\sin \,\beta \cos \,\alpha \,\cos \beta -{{\cos }^{2}}\,\alpha \,{{\sin }^{2}}\,\beta \\ & ={{\sin }^{2}}\,\alpha \,{{\cos }^{2}}\,\beta -{{\cos }^{2}}\,\alpha \,{{\sin }^{2}}\,\beta \\ & =\left( 1-{{\cos }^{2}}\alpha \right){{\cos }^{2}}\,\beta -{{\cos }^{2}}\,\alpha \,{{\sin }^{2}}\,\beta \end{align} \begin{align} & ={{\cos }^{2}}\,\beta -{{\cos }^{2}}\alpha {{\cos }^{2}}\,\beta -{{\cos }^{2}}\,\alpha \,{{\sin }^{2}}\,\beta \\ & ={{\cos }^{2}}\,\beta -{{\cos }^{2}}\alpha \left( {{\cos }^{2}}\,\beta +{{\sin }^{2}}\,\beta \right) \\ & ={{\cos }^{2}}\,\beta -{{\cos }^{2}}\alpha \left( 1 \right) \\ & ={{\cos }^{2}}\,\beta -{{\cos }^{2}}\alpha \\ \end{align} Hence, the left side of the given expression is equal to the right side, which is $\sin \left( \alpha +\beta \right)\sin \left( \alpha -\beta \right)={{\cos }^{2}}\,\beta -{{\cos }^{2}}\,\alpha$.

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