There are two basic approaches to trigonometry: the right triangle approach (prior section)
and the unit circle approach (this section).
Both approaches were introduced in Introduction to Trigonometry.
Be sure to read this prior section, since it covers important notation and conventions.

The sketch below shows two ways to get to the same terminal point: $30^\circ\,$ (positive angles are swept out counterclockwise; up) $330^\circ\,$ (negative angles are swept out clockwise; down) It doesn't matter how we get there! All that matters are the coordinates of the terminal point: $\displaystyle\cos 30^\circ = \cos(330^\circ) = \frac{\sqrt 3}{2}$ $\displaystyle \sin 30^\circ = \sin(330^\circ) = \frac{1}{2}$ 
What is the terminal point for $\,90^\circ\,$? From this information, find (where possible) the sine, cosine, and tangent of $\,90^\circ\,$. SOLUTION: As shown at right, the terminal point for $\,90^\circ\,$ is $\,(0,1)\,$.
Since division by zero is not allowed, $\,\tan 90^\circ\,$ is not defined. 
Let $\,P(x,y)\,$ be a point on the unit circle. Let $\,\theta\,$ be any angle that has $\,P\,$ as its terminal point.
Since $\,P(x,y)\,$ is on the unit circle $\,x^2 + y^2 = 1\,$, we have: $$\begin{gather} x^2 + (\frac 13)^2 = 1\cr\cr x^2 + \frac 19 = 1\cr\cr x^2 = 1  \frac 19 = \frac 89\cr\cr x = \pm\sqrt{\frac 89} = \pm \frac{\sqrt{4\cdot 2}}{3} = \pm \frac{2\sqrt{2}}{3} \end{gather} $$ Thus, $\displaystyle\,P(x,y) = P\bigl(\frac{2\sqrt{2}}{3},\frac 13\bigr)\,$.

On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
