Answer
(a) see proof below.
(b) $\frac{1}{2},-\frac{\sqrt 3}{2}, -\frac{\sqrt 3}{3}$
Work Step by Step
(a) To show that P is on the unit circle, plug in the coordinates of P into the equation $x^2+y^2=1$,
we see that $(-\frac{\sqrt 3}{2})^2+(\frac{1}{2})^2=\frac{3}{4}+\frac{1}{4}=1$, thus point P is on the unit circle.
(b) Based on the signs of the coordinates of P, we can see that P is in Quadrant II, thus we have
$sin(t)=y=\frac{1}{2},cos(t)=x=-\frac{\sqrt 3}{2}, tan(t)=\frac{y}{x}=-\frac{\sqrt 3}{3}$
