Find the value of $cos( heta)$ on the unit circle for a given $ heta$ and determine the exact coordinates of the corresponding point.

Let’s consider the angle $ \theta = \frac{7\pi}{6}$.

First, we determine the reference angle. Since $\frac{7\pi}{6}$ is in the third quadrant, we find the reference angle by subtracting $\pi$:

$ \theta_{ref} = \frac{7\pi}{6} – \pi = \frac{7\pi}{6} – \frac{6\pi}{6} = \frac{\pi}{6} $

The cosine of the reference angle $\frac{\pi}{6}$ is $\frac{\sqrt{3}}{2}$, but since we are in the third quadrant, the cosine value is negative:

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

The exact coordinates of the point on the unit circle corresponding to $\theta = \frac{7\pi}{6}$ are:

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

Let’s consider the angle $ heta = frac{10pi}{3}$.

First, we reduce the angle to within one full rotation of $2pi$:

$ heta = frac{10pi}{3} – 2pi = frac{10pi}{3} – frac{6pi}{3} = frac{4pi}{3} $

Now, $frac{4pi}{3}$ lies in the third quadrant. The reference angle is $frac{4pi}{3} – pi$:

$ heta_{ref} = frac{4pi}{3} – pi = frac{4pi}{3} – frac{3pi}{3} = frac{pi}{3} $

The cosine of the reference angle $frac{pi}{3}$ is $frac{1}{2}$, but since we are in the third quadrant, the cosine value is negative:

$ cosleft(frac{4pi}{3}
ight) = -frac{1}{2} $

Therefore, the exact coordinates of the point on the unit circle corresponding to $ heta = frac{10pi}{3}$ are:

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

Let’s consider the angle $ heta = frac{7pi}{4}$.

First, $frac{7pi}{4}$ lies in the fourth quadrant. The reference angle is $2pi – frac{7pi}{4}$:

$ heta_{ref} = 2pi – frac{7pi}{4} = frac{8pi}{4} – frac{7pi}{4} = frac{pi}{4} $

The cosine of the reference angle $frac{pi}{4}$ is $frac{sqrt{2}}{2}$, and since we are in the fourth quadrant, the cosine value remains positive:

$ cosleft(frac{7pi}{4}
ight) = frac{sqrt{2}}{2} $

Therefore, the exact coordinates of the point on the unit circle corresponding to $ heta = frac{7pi}{4}$ are:

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

Start Using PopAi Today