Find the coordinates of the point on the unit circle at which the angle is $ frac{7pi}{6} $

To find the coordinates of the point on the unit circle at which the angle is $ \frac{7\pi}{6} $, we use the following:

The unit circle has the equation:

$ x^2 + y^2 = 1 $

The coordinates of a point on the unit circle are given by:

$ (\cos(\theta), \sin(\theta)) $

For $ \theta = \frac{7\pi}{6} $:

$ \cos(\frac{7\pi}{6}) = -\frac{\sqrt{3}}{2} $

$ \sin(\frac{7\pi}{6}) = -\frac{1}{2} $

Therefore, the coordinates are:

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

The coordinates of the point on the unit circle at the angle $ frac{7pi}{6} $ can be found using:

$ (cos( heta), sin( heta)) $

For $ heta = frac{7pi}{6} $:

$ cos(frac{7pi}{6}) = -frac{sqrt{3}}{2} $

$ sin(frac{7pi}{6}) = -frac{1}{2} $

The coordinates are:

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

The coordinates of the point on the unit circle at the angle $ frac{7pi}{6} $ are:

$ cos(frac{7pi}{6}) = -frac{sqrt{3}}{2} $

$ sin(frac{7pi}{6}) = -frac{1}{2} $

Therefore:

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

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