Answer
Given point $(\frac{1}{2}, \frac{-\sqrt 3}{2})$ lies on the graph of unit circle.
Work Step by Step
Standard equation of unit circle is-
$x^{2} + y^{2}$ = 1
Coordinates of any point, lying on it will satisfy this equation.
Hence for given point $(\frac{1}{2}, \frac{-\sqrt 3}{2})$, substituting values of x and y in LS of standard equation of unit circle-
LS = $x^{2} + y^{2}$
= $(\frac{1}{2})^{2} + ( \frac{-\sqrt 3}{2})^{2}$
= $\frac{1}{4}+ \frac{3}{4}$
= $\frac{1 + 3}{4}$ = $\frac{4}{4}$ = 1 = RS
Hence given point $(\frac{1}{2}, \frac{-\sqrt 3}{2})$ lies on the graph of unit circle.
