Answer: (a) \(a=4\), \(b=\frac18\), \(c=7\). (b) \((12\pi,3)\).

Answer: (a) \(a=4\), \(b=\frac18\), \(c=7\). (b) \((12\pi,3)\).

The period of \(\sin bx\) is \(\frac{2\pi}{b}\). Since the period is \(16\pi\),

\(\displaystyle \frac{2\pi}{b}=16\pi\),

so \(b=\frac18\).

Use the point \((4\pi,11)\):

\(11=a\sin\left(\frac18\cdot4\pi\right)+c=a\sin\frac{\pi}{2}+c=a+c\).

Use the point \(\left(-\frac{4\pi}{3},5\right)\):

\(5=a\sin\left(-\frac{\pi}{6}\right)+c=-\frac12a+c\).

Solving

\(a+c=11\quad\text{and}\quad -\frac12a+c=5\)

gives \(a=4\) and \(c=7\).

Thus the curve is

\(y=4\sin\frac{x}{8}+7\).

The minimum value occurs when \(\sin\frac{x}{8}=-1\). In the interval \(0\le x\le16\pi\), this happens when

\(\frac{x}{8}=\frac{3\pi}{2}\),

so \(x=12\pi\). The corresponding \(y\)-value is

\(7-4=3\).

Therefore the minimum point is \((12\pi,3)\).