Remember: $$\cos x=1\Longleftrightarrow x=2k\pi\,,\,k\,\,\text{an integer}$$so$$\cos\left(2t+\frac{\pi}{3}\right)=1\Longleftrightarrow 2t+\frac{\pi}{3}=2k\pi$$
Assuming, as surely is the case, that it must be $\,t\geq 0\,$ , we get that$$2k\pi>\frac{\pi}{3}\Longrightarrow k=1,2,3,...$$and we can choose $k=1\Longrightarrow 2t+\frac{\pi}{3}=2\pi$
Amplitude, period and phase shift of can be recovered from the graph by noticing the coordinates of the peaks and troughs of the wave.
Let $y_{peak}$ and $y_{trough}$ denote the $y$-coordinates of the peaks and troughs of the wave. Then for the amplitude we have
$$ A=\frac{1}{2}\left(y_{peak}-y_{trough}\right) $$
From your graph there is no indication of the vertical scale, but let us suppose that the horizontal dashed lines are one unit apart. Then we would have $y_{peak}=2$ and $y_{trough}=-6$. This would give $A=\frac{1}{2}(2-(-6))=4$.
We can also use $y_{peak}$ and $y_{trough}$ to find the vertical shift $D$.
$$ D=\frac{1}{2}\left(y_{peak}+y_{trough}\right) $$
So, for your example, $D=\frac{1}{2}(2+(-6))=-2$.
This leaves the values of $B$ and $C$. But first, we must find the period $P$, which is straightforward.
The period P is the horizontal distance between two successive peaks of the graph. For your graph this would be a distance $P=3\pi$.
The value of $B$ is then found from
$$ B=\frac{2\pi}{P} $$
For your graph, then, we have $B=\frac{2\pi}{3\pi}=\frac{2}{3}$.
Finally we have the phase shift $\phi$.
The phase shift for the sine and cosine are computed differently. The easiest to determine from the graph is the phase shift of the cosine.
Let $x_{peak}$ denote the $x$-coordinate of the peak closest to the vertical axis. For your graph, we have $x_{peak}=0$.
$$\phi=x_{peak}\text{ for the cosine graph}$$
$$\phi=\left(x_{peak}-\frac{P}{4}\right)\text{ for the sine graph}$$
For your graph this gives phase shift $\phi=0$ for the cosine graph and phase shift $\phi=0-\frac{3\pi}{4}=-\frac{3\pi}{4}$ for the sine graph.
But for your equations, we need the value of $C$. The value of $C$ is found from the values of $\phi$ and $B$.
The equation for $C$ in terms of the phase shift $\phi$ is
$$ C=B\phi $$
So if we use the cosine function to model your graph we have $C=0$ and for the sine graph we have $C=\frac{2}{3}\cdot\left(-\frac{3\pi}{4}\right)=-\frac{\pi}{2}$.
So we have for both sine and cosine
- $A=4$
- $D=-2$
- $B=\frac{2}{3}$
For cosine, $C=0$ and for sine, $C=-\frac{\pi}{2}$.
Thus your graph can be represented by either of the two equations
$$ y=4\cos\left(\frac{2}{3}x \right)-2 $$
$$ y=4\sin\left(\frac{2}{3}x+\frac{\pi}{2}\right)-2 $$
Best Answer
Your first equation describes a sine function plus a constant. If $k\ne0$ there is no way to find a relation between the parameters of the two function because they are different. On the other hand if $k=0$ the relation is trivial: $$ y\mapsto x, x\mapsto t, a\mapsto A, b\mapsto\omega,h\mapsto -\phi/\omega. $$
Thus, for $k=0$ the functions are equivalent up to definition of parameters.