Find the coordinates of a point on the unit circle where the x-coordinate is $frac{1}{2}$.

The equation of the unit circle is given by:

$x^2 + y^2 = 1$

We are given that the x-coordinate is $\frac{1}{2}$. Substituting $x = \frac{1}{2}$ into the equation:

$\left(\frac{1}{2}\right)^2 + y^2 = 1$

$\frac{1}{4} + y^2 = 1$

Subtract $\frac{1}{4}$ from both sides:

$y^2 = 1 – \frac{1}{4}$

$y^2 = \frac{3}{4}$

Taking the square root of both sides:

$y = \pm \sqrt{\frac{3}{4}}$

$y = \pm \frac{\sqrt{3}}{2}$

Thus, the coordinates are:

$(\frac{1}{2}, \frac{\sqrt{3}}{2})$ and $(\frac{1}{2}, -\frac{\sqrt{3}}{2})$

The unit circle equation is:

$x^2 + y^2 = 1$

Given $x = frac{1}{2}$, we substitute it into the equation:

$left(frac{1}{2} ight)^2 + y^2 = 1$

$frac{1}{4} + y^2 = 1$

Isolate $y^2$:

$y^2 = 1 – frac{1}{4}$

$y^2 = frac{3}{4}$

Solve for $y$:

$y = pm frac{sqrt{3}}{2}$

Thus, the points are:

$(frac{1}{2}, frac{sqrt{3}}{2})$ and $(frac{1}{2}, -frac{sqrt{3}}{2})$

For the unit circle $x^2 + y^2 = 1$ with $x = frac{1}{2}$,

Substitute to get:

$left(frac{1}{2} ight)^2 + y^2 = 1$

$frac{1}{4} + y^2 = 1$

$y^2 = frac{3}{4}$

Thus, $y = pm frac{sqrt{3}}{2}$, and the coordinates are $(frac{1}{2}, frac{sqrt{3}}{2})$ and $(frac{1}{2}, -frac{sqrt{3}}{2})$.

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