Multiple choice - What sine function represents an amplitude of 4, a period of pi/2, no horizontal shift, and a vertical shift of −3?

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Multiple choice - What sine function represents an amplitude of 4, a period of pi/2, no horizontal shift, and a vertical shift of −3?

Mathematics
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a)f(x) = -3 sin 4x +4 b) f(x)= 4 sin 4x -3 c) f(x) = 4 sin (pi/2) x -3 d) f(x) = -3 sin (pi/2) x + 4
For the function \[\large a ~sin (bx-c)+d\] \[\large amplitude= \left| a \right|\]\[\large period=\frac{2\pi}{b}\]\[\large vertical ~shift = d\]
I'm not exactly sure how to plug in the inside of the parentheses. So far I have \[\left| 4 \right| \sin (???) - 3\] @LegendarySadist

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Other answers:

You can disregard the c in this example, as it's used for the phase shift, aka horizontal shift. So the b will be the integer before the \(x\)
I knew that c was the horizontal shift, but what exactly is b?
b is just the number before the x. For option #2, b is 4 and for option #3 b is \(\large \frac{\pi}{2}\)
I can see that b is before x, but I don't understand what I should plug in for b, or how to figure that out.
So we are looking for b, and we know the period. We'd set it up like this. \[\large \frac{2\pi}{b}=\frac{\pi}{2}\] Now cross multiply and solve for b
so b = 2?
\[\large \frac{2\pi}{2} = \frac{\pi}{2}\\\large \pi = \frac{\pi}{2}?\] No b is not 2
Herp, I forgot to multiply 2pi by pi!
\[\large \frac{2\pi}{b} = \frac{\pi}{2}\] Cross multiplying will get you \[\large 4\pi=\pi b\]
Can't you then just cancel out the pi's and get 4 = b?
Yep, that would be it. Good job :)
Thank you very much!
\[\huge \color{aqua}N\color{fuchsia}o \space \color{lime}P \color{orange}r \color{blue}o \color{maroon}b \color{red}l \color{olive}e \color{purple}m \ddot\smile \]
c:

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