1. Right Triangles
Consider the following equilateral triangle, with each side having a value of 2.
Drop a perpendicular segment from the top vertex to the base side.
The base side has been bisected into two segments of length 1.

2. Since the original triangle was equilateral, the base angles are 60° each.
Since the angle at the top of the triangle has been bisected, the original 60° angle has been split into new angles that are 30° each.

3. Consider the new triangle formed on the left side.
This is a right triangle.
We have just proven that in a triangle the short leg (always opposite the 30° angle) is always half the length of the hypotenuse.

4. Name the long leg b and determine its value.

5. This right triangle can give us the trigonometric function values of 30° and 60°. We first do the 30° angle.

6. We now do the 60° angle.

7. This leads us to some important values on the unit circle.
Recall that on the unit circle we have …

8. Consider the point (a,b) on the 30° ray of a unit circle.
Since (a,b) = (cos 30°, sin 30°) , we have

9. In radian form it would be …

10. Moving around the unit circle with reference angles of π/6 we have …

11. Example 1:
Find cos 5π/6
Since cos x is equal to the first coordinate of the point we have …

12. Example 2:
Find sin 7π/6
Since sin x is equal to the second coordinate of the point we have …

13. Example 3:
Find tan (-7π/6)
Since tan x is equal to b/a we have …

14. In similar fashion, moving around the unit circle with reference angles of π/3 (60°) we have …

15. Example 1:
Find cos 4π/3
Since cos x is equal to the first coordinate of the point we have …

16. Example 2:
Find sin -2π/3
Since sin x is equal to the second coordinate of the point we have …

17. Example 3:
Find tan (2π/3)
Since tan x is equal to b/a we have …