Answer
(a) $\frac{\pi}{3}$
(b) $P(\frac{1}{2},-\frac{\sqrt 3}{2})$
(c) $sin(t)=-\frac{\sqrt 3}{2},cos(t)=\frac{1}{2},tan(t)=-\sqrt 3, cot(t)=-\frac{\sqrt 3}{3}, sec(t)=2, csc(t)=-\frac{2\sqrt 3}{3}$
Work Step by Step
(a) It can be seen that $t=\frac{5\pi}{3}$ is in Quadrant IV, and we can find the reference number as $\bar t=2\pi-t=\frac{\pi}{3}$
(b) Given $\bar t=\frac{\pi}{3}$ in Quadrant IV, we have $P(\frac{1}{2},-\frac{\sqrt 3}{2})$
(c) With the results from above, we can obtain $sin(t)=y=-\frac{\sqrt 3}{2},cos(t)=x=\frac{1}{2},tan(t)=\frac{y}{x}=-\sqrt 3, cot(t)=\frac{x}{y}=-\frac{\sqrt 3}{3}, sec(t)=\frac{1}{x}=2, csc(t)=\frac{1}{y}=-\frac{2\sqrt 3}{3}$
