Find the coordinates of a point on the unit circle where the angle in standard position is given by $frac{7pi}{6}$ radians.
Answer 1
Emily Hall
To solve this, recall that an angle of $\frac{7\pi}{6}$ radians is in the third quadrant.
The reference angle for $\frac{7\pi}{6}$ is $\frac{7\pi}{6} – \pi = \frac{\pi}{6}$.
The coordinates for $\frac{\pi}{6}$ on the unit circle are $\left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)$.
Because $\frac{7\pi}{6}$ is in the third quadrant, both coordinates are negative.
Thus, the coordinates are $\left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right)$.
Answer 2
Ella Lewis
To solve this, we start by noting that an angle of $frac{7pi}{6}$ radians is in the third quadrant.
The reference angle for $frac{7pi}{6}$ is $frac{7pi}{6} – pi$, which simplifies to $frac{pi}{6}$.
For $ heta = frac{pi}{6}$, the coordinates on the unit circle are $left( cos frac{pi}{6}, sin frac{pi}{6}
ight)$, which are $left( frac{sqrt{3}}{2}, frac{1}{2}
ight)$.
Since $frac{7pi}{6}$ is in the third quadrant, we have:
$cos frac{7pi}{6} = -frac{sqrt{3}}{2}$
$sin frac{7pi}{6} = -frac{1}{2}$
Therefore, the coordinates are $left( -frac{sqrt{3}}{2}, -frac{1}{2}
ight)$.
Answer 3
Lily Perez
The angle $frac{7pi}{6}$ is in the third quadrant with a reference angle of $frac{pi}{6}$.
Coordinates at $frac{pi}{6}$ are $left( frac{sqrt{3}}{2}, frac{1}{2}
ight)$.
Thus, the coordinates at $frac{7pi}{6}$ are $left( -frac{sqrt{3}}{2}, -frac{1}{2}
ight)$.
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