1. The problem is to determine the reference angle for $\theta = \frac{11\pi}{6}$ and find $\sin \theta$, $\cos \theta$, and $\tan \theta$.\n\n2. First, find the reference angle. Recall the full angle $\theta = 2\pi = \frac{12\pi}{6}$. The reference angle is the difference between $2\pi$ and $\frac{11\pi}{6}$:\n$$\text{Reference angle} = 2\pi - \frac{11\pi}{6} = \frac{12\pi}{6} - \frac{11\pi}{6} = \frac{\pi}{6}$$\n\n3. Find $\sin \theta$: Since $\theta = \frac{11\pi}{6}$ is in the fourth quadrant, where sine is negative,\n$$\sin \frac{11\pi}{6} = -\sin \frac{\pi}{6} = -\frac{1}{2}$$\n\n4. Find $\cos \theta$: Cosine is positive in the fourth quadrant, so\n$$\cos \frac{11\pi}{6} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$$\n\n5. Find $\tan \theta$: Tangent in the fourth quadrant is negative. Using the reference angle,\n$$\tan \frac{11\pi}{6} = -\tan \frac{\pi}{6} = -\frac{1}{\sqrt{3}}$$\n\n**Final answers:**\n- Reference angle: $\frac{\pi}{6}$\n- $\sin \frac{11\pi}{6} = -\frac{1}{2}$\n- $\cos \frac{11\pi}{6} = \frac{\sqrt{3}}{2}$\n- $\tan \frac{11\pi}{6} = -\frac{1}{\sqrt{3}}$