Precalculus (6th Edition) Blitzer

The rectangular coordinates of $\left( 6,{{180}^{\circ }} \right)$ are $\left( -6,0 \right)$.
The rectangular coordinates of the point are $\left( x,y \right)$. Now rewrite the rectangular coordinates in terms of polar coordinates, $x=r\cos \theta \ \text{ and }\ y=r\sin \theta$ ……(I) Substituting the values of $r\ \text{ and }\ \theta$ in (I) we get \begin{align} & x=6\cos {{180}^{\circ }} \\ & =6\times \left( -1 \right) \\ & =-6 \end{align} And, \begin{align} & y=6\sin {{180}^{\circ }} \\ & =6\times 0 \\ & =0 \end{align} Therefore, the rectangular coordinates of $\left( 6,{{180}^{\circ }} \right)$ are $\left( -6,0 \right)$.