Step 1

Find the sine function satisying the properties.

The sine function: \(\displaystyle{f{{\left({x}\right)}}}={A}{\sin{{\left({\frac{{{2}\pi}}{{{p}}}}\cdot{\left({x}-{d}\right)}\right)}}}+{c}\) has amplitude A, \(\displaystyle{A}={\frac{{{f}_{{{1}}}-{f}_{{{2}}}}}{{{2}}}}\) with the period p hr, d is the horizontal shift and c is the vertical shift. If \(\displaystyle{d}{>}{0}\) graph shifts towards right and if \(\displaystyle{d}{<}{0}\) graph shifts towards left, \(\displaystyle{d}={\frac{{{t}_{{{1}}}+{t}_{{{2}}}}}{{{2}}}}\). If \(\displaystyle{c}{>}{0}\) graph shifts upward and if \(\displaystyle{c}{<}{0}\) graph shifts downward, \(\displaystyle{c}={\frac{{{f}_{{{1}}}+{f}_{{{2}}}}}{{{2}}}}\).

Here \(\displaystyle{f}_{{{1}}}\) and \(\displaystyle{f}_{{{2}}}\) are the maximum and minimum value of function respectively, \(\displaystyle{t}_{{{1}}}\) and \(\displaystyle{t}_{{{2}}}\) are the points where function attains maximum and minimum value respectively.

Sine function with period of 12 hr with minimum value of 10 at \(\displaystyle{t}={3}{h}{r}\) and maximum value of 16 at \(\displaystyle{t}={15}{h}{r}.\)

We have \(\displaystyle{p}={12},\ {f}_{{{1}}}={4},\ {f}_{{{2}}}=-{4},\ {t}_{{{1}}}={3}\) and \(\displaystyle{t}_{{{2}}}={9}\)

Step 2

Calculate A, d and c.

\(\displaystyle{A}={\frac{{{4}-{\left(-{4}\right)}}}{{{2}}}}\)

\(\displaystyle={4}\)

\(\displaystyle{d}={\frac{{{3}+{9}}}{{{2}}}}\)

\(\displaystyle={6}\)

and

\(\displaystyle{c}={\frac{{{4}-{4}}}{{{2}}}}\)

\(\displaystyle={0}\)

Substitute the values in the function:

\(\displaystyle{f{{\left({x}\right)}}}={4}{\sin{{\left({\frac{{{2}\pi}}{{{12}}}}\cdot{\left({x}-{6}\right)}\right)}}}+{0}\)

\(\displaystyle={4}{\sin{{\left({\frac{{\pi}}{{{6}}}}\cdot{\left({x}-{6}\right)}\right)}}}\)