## Precalculus (6th Edition) Blitzer

${{\left( 3\cos \theta -4\sin \theta \right)}^{2}}+{{\left( 4\cos \theta +3\sin \theta \right)}^{2}}$ Now, factorize the above expression. By using the formulae ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ and ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ ,with $A=3\cos \theta$ and $B=4\sin \theta$ , and $A=4\cos \theta$ and $B=3\sin \theta$ , respectively, for the numeric expression. \begin{align} & {{\left( 3\cos \theta -4\sin \theta \right)}^{2}}+{{\left( 4\cos \theta +3\sin \theta \right)}^{2}}={{\left( 3\cos \theta \right)}^{2}}-2.3\cos \theta .4\sin \theta +{{\left( 4sin\theta \right)}^{2}}+{{\left( 4\cos \theta \right)}^{2}} \\ & +2.4\cos \theta .3\sin \theta +{{\left( 3sin\theta \right)}^{2}} \\ & =9{{\cos }^{2}}\theta -24\cos \theta \sin \theta +16{{\sin }^{2}}\theta +16{{\cos }^{2}}\theta \\ & +24\cos \theta \sin \theta +9{{\sin }^{2}}\theta \end{align} Regroup the terms in the above expression and factor and simply it: \begin{align} & 9{{\cos }^{2}}\theta -24\cos \theta \sin \theta +16{{\sin }^{2}}\theta +16{{\cos }^{2}}\theta \\ & +24\cos \theta \sin \theta +9{{\sin }^{2}}\theta \\ & =9{{\cos }^{2}}\theta +9{{\sin }^{2}}\theta +16{{\sin }^{2}}\theta +16{{\cos }^{2}}\theta \\ & =9\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)+16\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right) \end{align} By using the Pythagorean identity of trigonometry ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ , the above expression can be further simplified as: \begin{align} & 9\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)+16\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)=9\left( 1 \right)+16\left( 1 \right) \\ & =9+16 \\ & =25 \end{align} Thus, the left side of the expression is equal to the right side, which is ${{\left( 3\cos \theta -4\sin \theta \right)}^{2}}+{{\left( 4\cos \theta +3\sin \theta \right)}^{2}}=25$.