Question

Rewrite the cosine function f(x) = cos(1/3x - pi/6) as a translated sine function

Answer 1

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:

\[
f(x) = \sin\left(\frac{1}{3}x + \frac{\pi}{3}\right)
\]