Chapter 13 - Vector Geometry - 13.7 Cylindrical and Spherical Coordinates - Exercises - Page 699: 2

So, $(2, \pi/3,-8) \Longrightarrow (1, \sqrt 3,-8) $.

We have $$ r=2, \quad \theta =\pi/3, \quad z=-8.$$ Now, we have $$ x=r\cos \theta=2 \cos \pi/3 =1,$$ $$ y=r\sin \theta=2 \sin \pi/3 =\sqrt 3,$$ $$ z=z=-8.$$ So, $(2, \pi/3,-8) \Longrightarrow (1, \sqrt 3,-8) $.