Find the angle whose cosine is $-frac{2}{3}$ on the unit circle

Given that $\cos(\theta) = -\frac{2}{3}$, we need to find the angle $\theta$ on the unit circle.

Since cosine represents the x-coordinate on the unit circle, we look for the angle in the second and third quadrants where the cosine values are negative.

In the second quadrant, we have:

$ \theta = \pi – \arccos\left(\frac{2}{3}\right) $

In the third quadrant, we have:

$ \theta = \pi + \arccos\left(\frac{2}{3}\right) $

Therefore, the angles whose cosine is $-\frac{2}{3}$ are:

$ \theta = \pi – \arccos\left(\frac{2}{3}\right) \text{ and } \theta = \pi + \arccos\left(\frac{2}{3}\right) $

Identify the angle $ heta$ such that $cos( heta) = -frac{2}{3}$ on the unit circle. The cosine function is negative in the second and third quadrants.

In the second quadrant, the angle is:

$ heta = pi – arccosleft(frac{2}{3}
ight) $

In the third quadrant, the angle is:

$ heta = pi + arccosleft(frac{2}{3}
ight) $

Thus, the solutions are:

$ heta = pi – arccosleft(frac{2}{3}
ight) ext{ and } heta = pi + arccosleft(frac{2}{3}
ight) $

Find $ heta$ such that $cos( heta) = -frac{2}{3}$.

In the second quadrant:

$ heta = pi – arccosleft(frac{2}{3}
ight) $

In the third quadrant:

$ heta = pi + arccosleft(frac{2}{3}
ight) $

Solutions:

$ heta = pi – arccosleft(frac{2}{3}
ight) ext{ and } heta = pi + arccosleft(frac{2}{3}
ight) $

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