Determine Specific Coordinates on the Unit Circle

Answer 1

Given a point on the unit circle defined by the angle $\theta$ in radians, find the coordinates of the point where $\theta = \frac{5\pi}{6}$.

First, recall that any point on the unit circle is defined by $(\cos \theta, \sin \theta)$. Here $\theta = \frac{5\pi}{6}$.

So, we need to find $\cos \left(\frac{5\pi}{6}\right)$ and $\sin \left(\frac{5\pi}{6}\right)$:

For $\cos \left(\frac{5\pi}{6}\right)$: Since $\frac{5\pi}{6}$ is in the second quadrant, $\cos \left(\frac{5\pi}{6}\right) = -\cos \left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$
For $\sin \left(\frac{5\pi}{6}\right)$: $\sin$ is positive in the second quadrant, so $\sin \left(\frac{5\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$.

Thus, the coordinates for $\theta = \frac{5\pi}{6}$ are $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right).$

Answer 2

Consider a point on the unit circle at an angle of $ heta = frac{7pi}{4}$ radians. Find its coordinates.

The coordinates of a point on the unit circle are given by $(cos heta, sin heta)$. Here $ heta = frac{7pi}{4}$:

For $cos left(frac{7pi}{4}
ight)$: Since $frac{7pi}{4}$ is in the fourth quadrant, $cos left(frac{7pi}{4}
ight) = cos left(-frac{pi}{4}
ight) = frac{1}{sqrt{2}} = frac{sqrt{2}}{2}$
For $sin left(frac{7pi}{4}
ight)$: Since $frac{7pi}{4}$ is in the fourth quadrant, $sin left(frac{7pi}{4}
ight) = sin left(-frac{pi}{4}
ight) = -frac{1}{sqrt{2}} = -frac{sqrt{2}}{2}$

Thus, the coordinates for $ heta = frac{7pi}{4}$ are $left(frac{sqrt{2}}{2}, -frac{sqrt{2}}{2}
ight).$

Answer 3

Find the coordinates on the unit circle where $ heta = frac{3pi}{4}$.

The coordinates are $(cos left(frac{3pi}{4}
ight), sin left(frac{3pi}{4}
ight))$:

The coordinates are $left(-frac{sqrt{2}}{2}, frac{sqrt{2}}{2}
ight).$

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