Graph the following function: $y = \sin \frac{1}{2}\left(x-\frac{\pi}{6}\right)$ I am working my way through Trigonometry by Gelfand and Saul. I am trying to work out the following question on p 183:

Graph the following function:
  $$y = \sin \frac{1}{2}\left(x-\frac{\pi}{6}\right)$$

I would like to check my understanding is correct. 
Firstly the authors talk about the importance of ensuring the equation is in the standard form of $y = a \sin k (x - \beta)$. I think $y = \sin \frac{1}{2}\left(x-\frac{\pi}{6}\right)$ is already in this standard form. 
Secondly I have drawn the graph (it is the lower one). Is it correct? Sorry it is hand drawn, I tried using WolframAlpha, but it would not let me format the x-axis using radians how I wanted.

Finally, does 
$$\sin \frac{1}{2}\left(x-\frac{\pi}{6}\right) = \sin\left(\frac{x}{2} -\frac{\pi}{12}\right)$$ so that $y = \sin \frac{1}{2}\left(x-\frac{\pi}{6}\right)$ is equivalent to writing  $y =\sin\left(\frac{x}{2} -\frac{\pi}{12}\right)$?
 A: Nice work, on your graph. You proceeded nicely (and your graph is very legible: I wish more of my students took the time to explain their work and draw as carefully as did you).
With respect to your last question: yes, $$y = \sin \left(\frac 12(x - \frac \pi 6)\right)\iff y = \sin\left(\frac x2 - \frac{\pi}{12}\right)$$ But note, as your author does, that you can only use the procedure you used above (when the equation is in standard form), to readily "read off" the what the period of the function is, and how much it is "shifted" to the right (or left, in other cases).
ADDED: On your second graph, I'd suggest adding a "tick-mark" on the y-axis to show the amplitude/magnitude of the function, as you did in the first graph.
A: One point of confusion might be the part where you have $(x-\frac{\pi}{6})$. Why is the function then shifted to the right instead of to  the left(as the minus sign suggests)? Actually $(x-\frac{\pi}{6})$ should be understood as shifting the coordinate system horizontally $\frac{\pi}{6}$ units to the left. Meaning that you just leave your graph as it is, and draw another coordinate system beneath it.
