Find the angle on the unit circle corresponding to the coordinates $left(-frac{2}{3}, y
ight)$.

To find the angle on the unit circle corresponding to the coordinates $\left(-\frac{2}{3}, y\right)$, we need to use the Pythagorean identity:

$x^2 + y^2 = 1$

Since $x = -\frac{2}{3}$, we plug this value into the equation:

$\left(-\frac{2}{3}\right)^2 + y^2 = 1$

$\frac{4}{9} + y^2 = 1$

Subtract $\frac{4}{9}$ from both sides:

$y^2 = 1 – \frac{4}{9}$

$y^2 = \frac{9}{9} – \frac{4}{9}$

$y^2 = \frac{5}{9}$

Take the square root of both sides:

$y = \pm\sqrt{\frac{5}{9}}$

$y = \pm\frac{\sqrt{5}}{3}$

The coordinates are $\left(-\frac{2}{3}, \pm\frac{\sqrt{5}}{3}\right)$.

We start by using the Pythagorean identity for the unit circle:

$x^2 + y^2 = 1$

Given $x = -frac{2}{3}$, we substitute it into the equation:

$left(-frac{2}{3}
ight)^2 + y^2 = 1$

$frac{4}{9} + y^2 = 1$

Subtract $frac{4}{9}$ from both sides:

$y^2 = 1 – frac{4}{9}$

$y^2 = frac{5}{9}$

Simplify $y$:

$y = pmfrac{sqrt{5}}{3}$

The points on the unit circle are $(-frac{2}{3}, frac{sqrt{5}}{3})$ and $(-frac{2}{3}, -frac{sqrt{5}}{3})$.

Use the identity $x^2 + y^2 = 1$:

$x = -frac{2}{3}$, so:

$left(-frac{2}{3}
ight)^2 + y^2 = 1$

$y^2 = frac{5}{9}$

$y = pmfrac{sqrt{5}}{3}$

Coordinates: $left(-frac{2}{3}, pmfrac{sqrt{5}}{3}
ight)$.

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