Answer
Yes, ($\frac{3}{4},-\frac{\sqrt 7}{4}$) is a point on the unit circle.
Work Step by Step
Since ($\frac{3}{4},-\frac{\sqrt 7}{4}$) is an (x, y) coordinate we can plug it into the equation of a circle, $x^{2}+y^{2}=1$.
$$(\frac{3}{4})^{2}+(-\frac{\sqrt 7}{4})^{2}=1$$ $$\frac{3\times3}{4\times4}+\frac{-\sqrt 7\times-\sqrt 7}{4\times4}=1$$ $$\frac{9}{16}+\frac{7}{16}=1$$ $$\frac{16}{16}=1$$ We end up with 1=1, which is true. Therefore, ($\frac{3}{4},-\frac{\sqrt 7}{4}$) is a point on the unit circle.