## Precalculus (6th Edition) Blitzer

The rectangular coordinates of $\left( 2,\frac{\pi }{6} \right)$ are $\left( \sqrt{3},1 \right)$.
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=2\cos \frac{\pi }{6} \\ & =2\times \frac{\sqrt{3}}{2} \\ & =\sqrt{3} \end{align} And \begin{align} & y=2\sin \frac{\pi }{6} \\ & =2\times \frac{1}{2} \\ & =1 \end{align} Therefore, the rectangular coordinates of $\left( 2,\frac{\pi }{6} \right)$ are $\left( \sqrt{3},1 \right)$.