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?

Note that $y=2 \sin \left( \frac{x}{2} \right)$ passes through the origin, and increases to its first peak at $x=\pi$.  
The function $2 \sin \left( \frac{x}{2} + b \right)$ is a horizontal shift of $2 \sin \left( \frac{x}{2} \right)$.  Since the picture shows a peak at $\frac{\pi}{2}$, we need to shift $y=2 \sin \left( \frac{x}{2} \right)$ to the left by $\frac{\pi}{2}$ to achieve this.
The function $2 \sin \left( \frac{x}{2} \right)$ shifted $\frac{\pi}{2}$ to the left is
$$
2 \sin \left( \frac{x+\frac{\pi}{2}}{2} \right) = 2 \sin \left( \frac{x}{2}+\frac{\pi}{4} \right)
$$
and so $b=\frac{\pi}{4}$.
