## Precalculus (6th Edition) Blitzer

The expression on the left side can be expanded by using the algebraic formulas ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$. Thus, the left side can be expressed as: \begin{align} & {{\left( \sin \theta +cos\theta \right)}^{2}}+{{\left( \sin \theta -cos\theta \right)}^{2}}=\left( {{\sin }^{2}}\theta +2\cos \theta .\sin \theta +{{\cos }^{2}}\theta \right)+ \\ & \left( {{\sin }^{2}}\theta -2\cos \theta .\sin \theta +{{\cos }^{2}}\theta \right) \\ & =\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right) \end{align} The expression can be further simplified by applying the Pythagorean identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ \begin{align} & \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)=1+1 \\ & =2 \end{align} Thus, the left side is equal to the right side ${{\left( \sin \theta +cos\theta \right)}^{2}}+{{\left( \sin \theta -cos\theta \right)}^{2}}=2$.