## Precalculus (6th Edition) Blitzer

Using the trigonometric identity $\sin \left( 2x \right)=2\sin x\cos x$ and ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ we will solve the provided identity. \begin{align} & \sin \left( 2x \right)=2\sin x\cos x+\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right) \\ & ={{\sin }^{2}}x+2\sin x\cos x+{{\cos }^{2}}x \end{align} Using the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we will solve the provided expression ${{\left( \sin x+\cos x \right)}^{2}}$ ${{\left( \sin x+\cos x \right)}^{2}}={{\sin }^{2}}x+2\sin x\cos x+{{\cos }^{2}}x$ Thus, $\sin (2x)+1={{\left( \sin x+\cos x \right)}^{2}}$ Hence, $\sin \left( 2x \right)+1={{\left( \sin x+\cos x \right)}^{2}}$.