To find $\sin\left(\frac{7\pi}{6}\right)$ using the unit circle, follow these steps:

1. Identify the reference angle:
   The angle $\frac{7\pi}{6}$ is in radians. First, subtract $\pi$ (or $\frac{6\pi}{6}$) from $\frac{7\pi}{6}$ to find the reference angle: $\frac{7\pi}{6} - \pi = \frac{7\pi}{6} - \frac{6\pi}{6} = \frac{\pi}{6}$ So, the reference angle is $\frac{\pi}{6}$.

2. Determine the quadrant:
   The angle $\frac{7\pi}{6}$ is slightly more than $\pi$, placing it in the third quadrant. In the third quadrant, the sine function is negative.

3. Find the sine of the reference angle:
   The reference angle is $\frac{\pi}{6}$, and from the unit circle, we know: $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

4. Apply the sign based on the quadrant:
   Since $\sin$ is negative in the third quadrant, the sine of $\frac{7\pi}{6}$ is: $\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}$

Final Answer:

$\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}$

Would you like further details or have any questions about this?

Here are five related questions:

1. What is the cosine of $\frac{7\pi}{6}$?
2. How do you find the tangent of $\frac{7\pi}{6}$?
3. What is the sine of $\frac{11\pi}{6}$?
4. How do the coordinates of points on the unit circle relate to sine and cosine values?
5. What is the general formula for sine in different quadrants?

Tip: In the unit circle, sine corresponds to the y-coordinate of the point on the circle.