## Precalculus (6th Edition) Blitzer

Let us consider the left side of the given expression: $\cos \left( \pi -x \right)$ By using the identity of trigonometry, $\cos \,\left( \alpha -\beta \right)=\cos \,\alpha \,\cos \beta +\sin \,\alpha \,\sin \,\beta$, the above expression can be further simplified as: \begin{align} & \cos \left( \pi -x \right)=\cos \,\pi \,\cos \,x+\sin \,\pi \,\sin \,x \\ & =-1\times \cos \,x+0\times \sin \,x \\ & =-\cos \,x+0 \\ & =-\cos \,x \end{align} Hence, the left side of the expression is equal to the right side, which is $\cos \left( \pi -x \right)=-\cos \,x$.