1. Stating the problem: Find the reference angle for $\theta = \frac{11\pi}{6}$, and compute $\sin \theta$, $\cos \theta$, and $\tan \theta$. 2. Reference angle: The reference angle is the acute angle that $\theta$ makes with the nearest x-axis. Since $\frac{11\pi}{6}$ lies in the fourth quadrant, the reference angle is $$\theta_{ref} = 2\pi - \frac{11\pi}{6} = \frac{12\pi}{6} - \frac{11\pi}{6} = \frac{\pi}{6}.$$ 3. Calculating $\sin \theta$: In the fourth quadrant, sine is negative. Using the reference angle, $$\sin\left(\frac{11\pi}{6}\right) = - \sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}.$$ 4. Calculating $\cos \theta$: Cosine is positive in the fourth quadrant, so $$\cos\left(\frac{11\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}.$$ 5. Calculating $\tan \theta$: Tangent is sine over cosine, so $$\tan\left(\frac{11\pi}{6}\right) = \frac{\sin\left(\frac{11\pi}{6}\right)}{\cos\left(\frac{11\pi}{6}\right)} = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}}.$$