I am doing a transformation problem of getting the graph of $\sin (2x – \pi/6)$ by applying transformations to $F(x) = \sin x$
In the process, I let $f(x) = F(2x) = \sin 2x$.
Next, I then let $g(x) = f(x – \pi/12) = \sin 2[x – \pi/12] = \sin (2x – \pi/6)$. The graphs are plotted as shown.
From f(x) to g(x), the above equations clearly shows there is a phase-shift of $\pi/12$ and this agrees with the red and blue lines
However, if I just comparing the functions $\sin (2x)$ and $\sin (2x – \pi/6)$ directly, shouldn’t there be just a right shift of $\pi/6$?.