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 relationship between sine and cosine:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
However, we must pay attention to the phase shift caused by the \( -\frac{\pi}{6} \) term.
Start by rewriting the cosine function:
\[ f(x) = \cos\left(\frac{1}{3}x - \frac{\pi}{6}\right) \]
Using the identity for cosine:
\[ f(x) = \sin\left(\left(\frac{1}{3}x - \frac{\pi}{6}\right) + \frac{\pi}{2}\right) \]
Now simplify the argument:
\[ \frac{1}{3}x - \frac{\pi}{6} + \frac{\pi}{2} = \frac{1}{3}x - \frac{\pi}{6} + \frac{3\pi}{6} = \frac{1}{3}x + \frac{2\pi}{6} = \frac{1}{3}x + \frac{\pi}{3} \]
So, we can write:
\[ f(x) = \sin\left(\frac{1}{3}x + \frac{\pi}{3}\right) \]
Thus, the function \( f(x) = \cos\left(\frac{1}{3}x - \frac{\pi}{6}\right) \) can be expressed as:
\[ f(x) = \sin\left(\frac{1}{3}x + \frac{\pi}{3}\right) \]
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