Given a point on the unit circle at an angle $ heta$ in radians, determine the exact coordinates and verify their correctness for $ heta = frac{7pi}{6}$.

We start with the unit circle formula:

$x^2 + y^2 = 1$

Given $\theta = \frac{7\pi}{6}$, we need to find the cosine and sine of this angle:

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

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

Thus, the coordinates are:

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

Verification:

$\left( -\frac{\sqrt{3}}{2} \right)^2 + \left( -\frac{1}{2} \right)^2 = \frac{3}{4} + \frac{1}{4} = 1$

The coordinates are correct.

Given $ heta = frac{7pi}{6}$, we identify the reference angle and the related trigonometric values:

$ heta = pi + frac{pi}{6}$

Therefore,

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

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

The coordinates are:

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

Check:

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

Hence, the given coordinates are correct.

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

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

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

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

Verification:

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

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