Answer:
[tex](\frac{\sqrt{3}}{2},-\frac{1}{2})[/tex]
Step-by-step explanation:
The unit circle has the parametric equation:
[tex]x=\cos \theta[/tex]
[tex]y=\sin \theta[/tex]
where [tex]\theta=\frac{11\pi }{6}[/tex] in our case is the terminal side of the angle in standard position.
We substitute to get:
[tex]x=\cos \frac{11\pi }{6}=\frac{\sqrt{3}}{2}[/tex]
[tex]y=\sin \frac{11\pi }{6}=-\frac{1}{2}[/tex]
[tex]\therefore (\frac{\sqrt{3}}{2},-\frac{1}{2})[/tex] is the required point.
