Question

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

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