Determine the coordinates on the unit circle for the angle whose cosine is $-frac{2}{3}$

To find the coordinates on the unit circle for the angle whose cosine is $-\frac{2}{3}$, we use the Pythagorean identity:

$\cos^2\theta + \sin^2\theta = 1$

Given that $\cos\theta = -\frac{2}{3}$, we substitute this into the identity:

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

This gives:

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

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

$\sin^2\theta = 1 – \frac{4}{9}$

$\sin^2\theta = \frac{9}{9} – \frac{4}{9}$

$\sin^2\theta = \frac{5}{9}$

Taking the square root of both sides, we get:

$\sin\theta = \pm\frac{\sqrt{5}}{3}$

So, the coordinates on the unit circle are:

$\left(-\frac{2}{3},\frac{\sqrt{5}}{3}\right) \text{and} \left(-\frac{2}{3},-\frac{\sqrt{5}}{3}\right)$

Given the cosine of an angle $ heta$ on the unit circle is $-frac{2}{3}$, we start from the Pythagorean identity:

$cos^2 heta + sin^2 heta = 1$

Substitute $cos heta = -frac{2}{3}$:

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

$frac{4}{9} + sin^2 heta = 1$

Isolate $sin^2 heta$ by subtracting $frac{4}{9}$ from 1:

$sin^2 heta = 1 – frac{4}{9}$

$sin^2 heta = frac{5}{9}$

Taking the square root, we get:

$sin heta = pmfrac{sqrt{5}}{3}$

Thus, the coordinates are:

$left(-frac{2}{3},frac{sqrt{5}}{3}
ight) ext{and} left(-frac{2}{3},-frac{sqrt{5}}{3}
ight)$

Start with $cos heta = -frac{2}{3}$:

$cos^2 heta + sin^2 heta = 1$

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

$frac{4}{9} + sin^2 heta = 1$

$sin^2 heta = frac{5}{9}$

$sin heta = pmfrac{sqrt{5}}{3}$

Coordinates:

$left(-frac{2}{3},frac{sqrt{5}}{3}
ight) ext{and} left(-frac{2}{3},-frac{sqrt{5}}{3}
ight)$

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