If $\theta=\frac{-11 \pi}{6}$, then
\[
\sin (\theta)=
\]
\[
\cos (\theta)=
\]

Step 1 ：The given angle is \(\theta = \frac{-11\pi}{6}\).

Step 2 ：The sine and cosine functions are periodic with a period of \(2\pi\). This means that \(\sin(\theta + 2\pi) = \sin(\theta)\) and \(\cos(\theta + 2\pi) = \cos(\theta)\) for any real number \(\theta\).

Step 3 ：So, if \(\theta = \frac{-11\pi}{6}\), we can add \(2\pi\) to \(\theta\) until we get a number in the range \([0, 2\pi)\). This will not change the values of \(\sin(\theta)\) and \(\cos(\theta)\).

Step 4 ：The equivalent angle in the range \([0, 2\pi)\) is \(\theta = 0.524\) (rounded to three decimal places).

Step 5 ：The sine of this angle is \(\sin(\theta) = 0.500\) (rounded to three decimal places) and the cosine of this angle is \(\cos(\theta) = 0.866\) (rounded to three decimal places).

Step 6 ：Therefore, we have: \(\sin\left(\frac{-11\pi}{6}\right) = \boxed{0.500}\) and \(\cos\left(\frac{-11\pi}{6}\right) = \boxed{0.866}\).