Find the coordinates of the point on the unit circle corresponding to the angle whose cosine is $-frac{2}{3}$.

Given the cosine of the angle is $-\frac{2}{3}$,

First, find the sine of the angle using the Pythagorean identity:

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

Substitute $\cos\theta = -\frac{2}{3}$:

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

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

$\sin^2\theta = 1 – \frac{4}{9}$

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

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

Taking the square root,

$\sin\theta = \pm\sqrt{\frac{5}{9}}$

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

Thus, the coordinates are:

$(-\frac{2}{3}, \frac{\sqrt{5}}{3})$ or $(-\frac{2}{3}, -\frac{\sqrt{5}}{3})$

To find the coordinates of the point on the unit circle where the cosine is $-frac{2}{3}$, we use the Pythagorean identity:

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

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

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

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

$sin^2 heta = 1 – frac{4}{9}$

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

Taking the square root:

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

Therefore, the coordinates are:

$(-frac{2}{3}, frac{sqrt{5}}{3})$ or $(-frac{2}{3}, -frac{sqrt{5}}{3})$

Given $cos heta = -frac{2}{3}$, use the identity:

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

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

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

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

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

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

Coordinates:

$(-frac{2}{3}, frac{sqrt{5}}{3})$ or $(-frac{2}{3}, -frac{sqrt{5}}{3})$

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