Trigonometric Function Modeling Question: In a normal respiratory cycle the volume of air that moves into and out of the lungs is about $500$ mL. The reserve and residue volumes of air that remain in the lungs occupy about $2000$ mL and a single respiratory cycle for an average human takes about $4$ seconds. Find a model for the total volume of air $V(t)$ in the lungs as a function of time.
Answer: $$500\sin\bigg(\frac{\pi}2t\bigg) + 2000$$
Am I correct? If not, then could you explain where I went wrong?
 A: I can spot one problem. It seems to me that the answer should be
$$500\sin\bigg(\frac{\pi t}{2}\bigg)+2500$$
Because the question makes it sound like there should be at least $2000$mL of air in the lungs at all times, and with your model, the amount of air may dip down to $1500$ when the sinusoidal part of your function hits a minimum of $-500$. But other than that, it looks good!
EDIT: As noted by @lulu, it's possible that the writer of this question may have intended the answer to be
$$250\sin\bigg(\frac{\pi t}{2}\bigg)+2250$$
because this allows for a $500$mL difference from $2000$mL in the lungs, rather than a $500$mL range above and one below $2000$mL.
A: With a period $4$ second we observe:
\begin{cases}
t=0 ; V(t)=2000,\\
t=1 ; ?,\\
t=2 ; V(t)=2500,\\
t=3 ; ?.
\end{cases}
for a regular respiratory cycle we guess the amount of volume in lung
\begin{cases}
t=0 ; V(0)=2000,\\
t=1 ; V(1)=2250,\\
t=2 ; V(2)=2500,\\
t=3 ; V(3)=2250.
\end{cases}
these points make a periodic model which can be as a Fourier series. So with discrete (time) Fourier series for $\{2000,2250,2500,2250\}$ we have from main topics that for $k$ observed points $\{z_1,z_2,z_3,\cdots,z_k\}$ in a model the series is
$$f(t)=\frac{a_0}{2}+\sum_{n=1}^{q}\Big(a_n\cos\frac{2n\pi}{k}t+b_n\sin\frac{2n\pi}{k}t\Big)$$
where $a_0$, $a_n$ and $b_n$ are:
\begin{eqnarray*}
a_0 &=& \frac{2}{k}\sum_{t=1}^kz_t                    \\
a_n &=& \frac{2}{k}\sum_{t=1}^kz_t\cos\frac{2n\pi}{k}t ~~~~~,~~~~~n=1,2,\cdots,q \\
b_n &=& \frac{2}{k}\sum_{t=1}^kz_t\sin\frac{2n\pi}{k}t ~~~~~,~~~~~n=1,2,\cdots,q
\end{eqnarray*}
and here $k=4$ and $q=\dfrac{k}{2}=2$ and $n=2$ so we find $a_1=a_2=b_1=0$ and $b_2=-250$. These coefficients make the series
$$\color{blue}{f(t)=2250-250\sin\left(\dfrac{\pi}{2}t\right)}$$
