There are two basic approaches to trigonometry:
Both approaches were introduced in Introduction to Trigonometry.

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: $$\begin{gather} \cssId{s60}{x^2 + (\frac 13)^2 = 1}\cr\cr \cssId{s61}{x^2 + \frac 19 = 1}\cr\cr \cssId{s62}{x^2 = 1  \frac 19 = \frac 89}\cr\cr \cssId{s63}{x = \pm\sqrt{\frac 89} = \pm \frac{\sqrt{4\cdot 2}}{3} = \pm \frac{2\sqrt{2}}{3}} \end{gather} $$ 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. 
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