Find the Cotangent of an Angle on the Unit Circle

To find the cotangent of an angle $\theta$ on the unit circle, we use the identity:

$ \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} $

Given $\theta = \frac{3\pi}{4}$, we know from the unit circle that:

$ \cos \left( \frac{3\pi}{4} \right) = -\frac{\sqrt{2}}{2} $

and

$ \sin \left( \frac{3\pi}{4} \right) = \frac{\sqrt{2}}{2} $

Therefore,

$ \cot \left( \frac{3\pi}{4} \right) = \frac{\cos \left( \frac{3\pi}{4} \right)}{\sin \left( \frac{3\pi}{4} \right)} = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1 $

So, the cotangent of $\frac{3\pi}{4}$ is $-1$.

We need to find the cotangent of the angle $ heta = frac{5pi}{3}$ using the unit circle:

From the unit circle, we have:

$ cos left( frac{5pi}{3}
ight) = frac{1}{2} $

and

$ sin left( frac{5pi}{3}
ight) = -frac{sqrt{3}}{2} $

Using the cotangent identity:

$ cot heta = frac{cos heta}{sin heta} $

We get:

$ cot left( frac{5pi}{3}
ight) = frac{cos left( frac{5pi}{3}
ight)}{sin left( frac{5pi}{3}
ight)} = frac{frac{1}{2}}{-frac{sqrt{3}}{2}} = -frac{1}{sqrt{3}} = -frac{sqrt{3}}{3} $

Thus, the cotangent of $frac{5pi}{3}$ is $-frac{sqrt{3}}{3}$.

To find the cotangent of $ heta = frac{pi}{6}$:

We know:

$ cos left( frac{pi}{6}
ight) = frac{sqrt{3}}{2} $

and

$ sin left( frac{pi}{6}
ight) = frac{1}{2} $

Thus:

$ cot left( frac{pi}{6}
ight) = frac{cos left( frac{pi}{6}
ight)}{sin left( frac{pi}{6}
ight)} = frac{frac{sqrt{3}}{2}}{frac{1}{2}} = sqrt{3} $

Therefore, the cotangent of $frac{pi}{6}$ is $sqrt{3}$.

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