## Trigonometry 7th Edition

$(\frac{\sqrt5}{3}, -\frac{2}{3})$) is found on the graph of the unit circle.
A unit circle is a circle with a radius of one, with a center on the origin (0,0). Therefore the point $(\frac{\sqrt5}{3}, -\frac{2}{3})$), is found on the unit circle. To check for this, substitute the x and y values of the point into the area of a circle as well as the radius. $x^2+y^2=r$ $(\frac{\sqrt5}{3})^2+(-\frac{2}{3})^2=1$ Substitute. $\frac{5}{9}+\frac{4}{9}=1$ $\frac{9}{9}=1$ $1=1$