Find the coordinates of the point on the unit circle corresponding to the angle $frac{7pi}{6}$

To find the coordinates on the unit circle for the angle $\frac{7\pi}{6}$, we use the unit circle properties:

The unit circle coordinates $(x, y)$ for an angle $\theta$ are $(\cos(\theta), \sin(\theta))$.

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

$ x = \cos\left(\frac{7\pi}{6}\right) $

$ y = \sin\left(\frac{7\pi}{6}\right) $

Using trigonometric identities:

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

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

Therefore, the coordinates are:

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

To find the coordinates on the unit circle for the angle $frac{7pi}{6}$:

The unit circle coordinates $(x, y)$ for an angle $ heta$ are $(cos( heta), sin( heta))$.

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

$ x = cosleft(frac{7pi}{6}
ight) = -frac{sqrt{3}}{2} $

$ y = sinleft(frac{7pi}{6}
ight) = -frac{1}{2} $

Thus, the coordinates are:

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

For the angle $frac{7pi}{6}$ on the unit circle:

Coordinates $(cos(frac{7pi}{6}), sin(frac{7pi}{6}))$:

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

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

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

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