# How do you find the point (x,y) on the unit circle that corresponds to the real number t=(11pi)/6?

($\frac{\sqrt{3}}{2}$, -$\frac{1}{2}$)
$\frac{11 \pi}{6}$ corresponds to 330 degrees on the unit circle, which also has the same point values as $\frac{\pi}{6}$, but the y value is negative because it is in the 4th quadrant.