Example 1 from Algebra and Trigonometry, 7.4 The Other Trigonometric Functions

Finding Trigonometric Functions from a Point on the Unit Circle

The point $(- \frac{\sqrt{3}}{2}, \frac{1}{2})$ is on the unit circle, as shown in Figure 2. Find $\sin t, \cos t, \tan t, \sec t, \csc t,$ and $\cot t .$

This is an image of a graph of circle with angle of t inscribed and with radius 1. Point of (negative square root of 3 over 2, 1/2) is at intersection of terminal side of angle and edge of circle.

Figure 2

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Because we know the $(x, y)$ coordinates of the point on the unit circle indicated by angle $t,$ we can use those coordinates to find the six functions:

$\begin{matrix} sin t & = y & = \frac{1}{2} \\ cos t & = x & = - \frac{\sqrt{3}}{2} \\ tan t & = \frac{y}{x} & = \frac{\frac{1}{2}}{- \frac{\sqrt{3}}{2}} & = \frac{1}{2} (- \frac{2}{\sqrt{3}}) & = - \frac{1}{\sqrt{3}} & = - \frac{\sqrt{3}}{3} \\ sec t & = \frac{1}{x} & = \frac{1}{- \frac{\sqrt{3}}{2}} & = - \frac{2}{\sqrt{3}} & = - \frac{2 \sqrt{3}}{3} \\ csc t & = \frac{1}{y} & = \frac{1}{\frac{1}{2}} & = 2 \\ cot t & = \frac{x}{y} & = \frac{- \frac{\sqrt{3}}{2}}{\frac{1}{2}} & = - \frac{\sqrt{3}}{2} (\frac{2}{1}) & = - \sqrt{3} \end{matrix}$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

The point $(\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2})$ is on the unit circle, as shown in Figure 3. Find $\sin t, \cos t, \tan t, \sec t, \csc t,$ and $\cot t .$

This is an image of a graph of circle with angle of t inscribed with radius 1. Point of (square root of 2 over 2, negative square root of 2 over 2) is at intersection of terminal side of angle and edge of circle.

Figure 3

How did it go?

Solution

Your turn

Solution

Your turn