To rewrite the cosine function \( f(x) = \cos\left(\frac{1}{3}x - \frac{\pi}{6}\right) \) as a translated sine function, we can use the identity that relates cosine and sine:
\[ \cos\left(\theta\right) = \sin\left(\frac{\pi}{2} - \theta\right) \]
First, we can set:
\[ \theta = \frac{1}{3}x - \frac{\pi}{6} \]
Then we can rewrite the cosine function:
\[ f(x) = \cos\left(\frac{1}{3}x - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{2} - \left(\frac{1}{3}x - \frac{\pi}{6}\right)\right) \]
Now, simplify the argument of the sine function:
\[ \frac{\pi}{2} - \left(\frac{1}{3}x - \frac{\pi}{6}\right) = \frac{\pi}{2} + \frac{\pi}{6} - \frac{1}{3}x \]
To combine \( \frac{\pi}{2} \) and \( \frac{\pi}{6} \), we first convert \( \frac{\pi}{2} \) into sixths:
\[ \frac{\pi}{2} = \frac{3\pi}{6} \]
So we have:
\[ \frac{3\pi}{6} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \]
Now we can substitute back into the sine function:
\[ f(x) = \sin\left(\frac{2\pi}{3} - \frac{1}{3}x\right) \]
Which can be rewritten by factoring out a negative sign:
\[ f(x) = \sin\left(-\left(\frac{1}{3}x - \frac{2\pi}{3}\right)\right) = -\sin\left(\frac{1}{3}x - \frac{2\pi}{3}\right) \]
To make it look like the options given, we can use the property:
\[ \sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b) \]
However, the options you've given are simpler. We can go on rewriting:
\[ f(x) = \sin\left(\frac{1}{3}x + \frac{\pi}{3}\right) \]
This can be checked against the provided options. Thus the answer is:
A. \( f(x) = \sin\left(\frac{1}{3}x + \frac{\pi}{3}\right) \)