Why does sin(x/2 - 2 pi) graph the same as sin(x/2)?

[vns]

I would expect the shift to right by unit of 2 pi to produce a different graph since the period of sin(x/2) is 4 pi?

 

[HallsofIvy]

But sine still has period 2pi: sin(x/2- 2pi)= sin(x/2)cos(2pi)- cos(x/2)sin(2pi)= sin(x/2)(1)- cos(x/2)(0)= sin(x/2).

If it were sin((x- 2pi)/2) THEN the graph would be different.

 

[vns]

Okay. I see. So, I'm going to say that, as a rule of thumb, it is best to do the horizontal shifting before the horizontal stretching in order to prevent confusion. Let me know if you see a problem with this reasoning.

Thanks for your help.

 

[pka]

Okay. I see. So, I'm going to say that, as a rule of thumb, it is best to do the horizontal shifting before the horizontal stretching in order to prevent confusion.

​It is certainly not a rule of thumb ! It is a result of the laws of the sine of a sum.
For any integer \(\displaystyle k\) we know that \(\displaystyle \sin(2k\pi)=0~\&~\cos(2k\pi)=1\).
We also know that \(\displaystyle \sin(A\pm B)=\sin(A)\cos(B)\pm\sin(B)\cos(A)\)
So \(\displaystyle \sin(A\pm 2k\pi)=\sin(A)\)

 

[vns]

Okay, I see. I understand the need to know the laws of sine. And, I plan to remember this. However, I know my brain likes to forget things. In this case, if I shift horizontally first and then expand horizontally, I get the right number. If I expand horizontally first and then shift, I will need to remember the rules stated above. I definitely plan to do this; however, it is possible to be in the middle of a test and forget. So, I wanted to come up with a backup plan. Don't always have time to do things the long way. Long way helps understand big picture, but I need shortcuts for tests, or I'm not able to finish them.

Thanks for all the explanations. This is exactly what I was looking for. If I've said something completely off-base, please correct me.