Since $\tan{\theta} = 2$ and $\theta$ is in the third quadrant, we have:

$$\sin{\theta} = 2\cos{\theta}$$

Recall the Pythagorean trigonometric identity:

$$\sin^2{\theta} + \cos^2{\theta} = 1$$

Substitute $\sin{\theta}$ with $2\cos{\theta}$:

$$(2\cos{\theta})^2 + \cos^2{\theta} = 1$$

Expand and simplify:

$$4\cos^2{\theta} + \cos^2{\theta} = 1$$

$$5\cos^2{\theta} = 1$$

$$\cos^2{\theta} = \frac{1}{5}$$

Since $\theta$ is in the third quadrant, $\cos{\theta}$ is negative:

$$\cos{\theta} = -\sqrt{\frac{1}{5}} = -\frac{\sqrt{5}}{5}$$

Recall the cofunction identity:

$$\sin({\frac{\pi}{2} - \theta}) = \cos{\theta}$$

Therefore:

$$\sin({\frac{\pi}{2} - \theta}) = -\frac{\sqrt{5}}{5}$$

\boxed{-\frac{\sqrt{5}}{5}}