Find the coordinates of the point on the unit circle where the inverse sine function is equal to $ frac{1}{2} $

To find the coordinates of the point on the unit circle where the inverse sine function, $ \sin^{-1}(x) $, is equal to $ \frac{1}{2} $:

We need to solve the equation:

$ \sin^{-1}(y) = \frac{1}{2} $

The angle whose sine is $ \frac{1}{2} $ is:

$ \theta = \frac{\pi}{6} $

On the unit circle, the coordinates corresponding to this angle are:

$ (\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6})) $

Evaluating the trigonometric functions, we get:

$ \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} $

$ \sin(\frac{\pi}{6}) = \frac{1}{2} $

So, the coordinates are:

$ \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) $

To find the coordinates of the point on the unit circle where the inverse sine function is equal to $ frac{1}{2} $:

Solve the equation:

$ sin^{-1}(y) = frac{1}{2} $

The angle whose sine is $ frac{1}{2} $ is:

$ heta = frac{pi}{6} $

On the unit circle, the coordinates are:

$ (cos(frac{pi}{6}), sin(frac{pi}{6})) $

Thus, we have:

$ left( frac{sqrt{3}}{2}, frac{1}{2}
ight) $

To find the coordinates of the point on the unit circle where $ sin^{-1}(x) = frac{1}{2} $:

The angle is:

$ frac{pi}{6} $

Coordinates:

$ left( frac{sqrt{3}}{2}, frac{1}{2}
ight) $

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