Trigonometric Functions Worksheet - Evaluate trigonometric expressions at specific angles.

To solve the trigonometric evaluation problems, we will use the unit circle and trigonometric identities. Let's go through each problem step by step.

Problem 1: Evaluate $\cos\left(\frac{\pi}{3}\right)$

- The angle $\frac{\pi}{3}$ corresponds to $60^\circ$ on the unit circle.
- From the unit circle, we know:
\[
\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}
\]

Problem 2: Evaluate $\sin\left(\frac{3\pi}{4}\right)$

- The angle $\frac{3\pi}{4}$ corresponds to $135^\circ$ in the second quadrant.
- In the second quadrant, sine is positive.
- The reference angle for $\frac{3\pi}{4}$ is $\pi - \frac{3\pi}{4} = \frac{\pi}{4}$.
- From the unit circle, we know:
\[
\sin\left(\frac{3\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}
\]

Problem 3: Evaluate $\tan\left(\frac{7\pi}{6}\right)$

- The angle $\frac{7\pi}{6}$ corresponds to $210^\circ$ in the third quadrant.
- In the third quadrant, tangent is positive.
- The reference angle for $\frac{7\pi}{6}$ is $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$.
- From the unit circle, we know:
\[
\tan\left(\frac{7\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}
\]

Problem 4: Evaluate $\sec\left(-\frac{\pi}{6}\right)$

- The angle $-\frac{\pi}{6}$ corresponds to $-30^\circ$, which is coterminal with $330^\circ$ in the fourth quadrant.
- In the fourth quadrant, secant is positive.
- The reference angle for $-\frac{\pi}{6}$ is $\frac{\pi}{6}$.
- From the unit circle, we know:
\[
\sec\left(-\frac{\pi}{6}\right) = \sec\left(\frac{\pi}{6}\right) = \frac{1}{\cos\left(\frac{\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}
\]

Problem 5: Evaluate $\csc\left(\frac{5\pi}{3}\right)$

- The angle $\frac{5\pi}{3}$ corresponds to $300^\circ$ in the fourth quadrant.
- In the fourth quadrant, cosecant is negative.
- The reference angle for $\frac{5\pi}{3}$ is $2\pi - \frac{5\pi}{3} = \frac{\pi}{3}$.
- From the unit circle, we know:
\[
\csc\left(\frac{5\pi}{3}\right) = \csc\left(\frac{\pi}{3}\right) = \frac{1}{\sin\left(\frac{\pi}{3}\right)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}
\]
Since it is in the fourth quadrant, we take the negative value:
\[
\csc\left(\frac{5\pi}{3}\right) = -\frac{2\sqrt{3}}{3}
\]

Problem 6: Evaluate $\cot\left(\frac{\pi}{4}\right)$

- The angle $\frac{\pi}{4}$ corresponds to $45^\circ$.
- From the unit circle, we know:
\[
\cot\left(\frac{\pi}{4}\right) = \frac{1}{\tan\left(\frac{\pi}{4}\right)} = \frac{1}{1} = 1
\]

Problem 7: Evaluate $\tan(2\pi)$

- The angle $2\pi$ corresponds to a full rotation, which is coterminal with $0^\circ$.
- From the unit circle, we know:
\[
\tan(2\pi) = \tan(0) = 0
\]

Problem 8: Evaluate $\sec\left(-\frac{5\pi}{6}\right)$

- The angle $-\frac{5\pi}{6}$ corresponds to $-150^\circ$, which is coterminal with $210^\circ$ in the third quadrant.
- In the third quadrant, secant is negative.
- The reference angle for $-\frac{5\pi}{6}$ is $\pi + \frac{\pi}{6} = \frac{7\pi}{6}$, but we use $\frac{\pi}{6}$ as the reference.
- From the unit circle, we know:
\[
\sec\left(-\frac{5\pi}{6}\right) = \sec\left(\frac{7\pi}{6}\right) = \frac{1}{\cos\left(\frac{7\pi}{6}\right)} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}
\]

Final Answer:

\[
\boxed{\frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}, -\frac{2\sqrt{3}}{3}, 1, 0, -\frac{2\sqrt{3}}{3}}
\]