<s>[INST] Find $b$ in $2\sin(\frac{x}{2}+b)$ given the graph Following is the graph of the function. $2\sin \left(\dfrac{x}{2}+b\right)$

I tried solving the equation using the point $\left (\dfrac{\pi}{2},2\right)$:
$$2 = 2\sin\left(\dfrac{x}{2}+b\right) \Leftrightarrow \\
1 = \sin\left(\dfrac{x}{2}+b\right)\Leftrightarrow \\
\frac{x}{2}+b = k\pi + (-1)^k\arcsin(1) \Leftrightarrow \\
b = 2k\pi$$
But my book $b = \dfrac{\pi}{4}$. How do I solve this?
 [/INST] $\sin\left( \dfrac x 2 + b \right) = 1$ when $\dfrac x 2 + b = \dfrac \pi 2,$ and that implies $x=\pi-2b.$
Since it appears that the point where the curve reaches its maximum is half-way between $0$ and $\pi/2,$ we have $\dfrac\pi 2 = \pi - 2b,$ and from that we get $b=\dfrac \pi 4.$
A bit of an optical illusion seems to afflict this graph: The positioning of the $\text{“}0\text{''}$ a bit to the right of the zero point, presumably to avoid putting it right on top of that vertical axis, makes it look as if that maximum point may be closer to $0$ than to $\pi/2.$ But closer inspection dispells that.
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