Trigonometry graphs sinusoidal waves i need help on this questions. I couldn't figure how to determine for both question A and B.
But i have the answers for them, i just don't understand how the amplitude is 3 and so on.

 A: To find the amplitude, take half the difference of the maximum and minimum values.  In this case, the amplitude is
$$a = \frac{1 - (-5)}{2} = \frac{6}{2} = 3$$
If you subtract the amplitude from the maximum value, you will find the average value.  In this case, the average value of the function is $1 - 3 = -2$.  If there were no vertical shift, the average value of the function would be zero.  Therefore, the vertical shift is 
$$d = 0 - (-2) = 2$$
Since the sine function assumes its average value at $0$, there is no phase shift.  Hence, $c = 0$.
Observe that the graph shows that $3/4$ of a period is $1/2$.  Hence, one period is 
$$\frac{\dfrac{1}{2}}{\dfrac{3}{4}} = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}$$
Since $\sin t$ has period $2\pi$, you can find the frequency by setting $bt = 2\pi$, where $t$ is the period of the sinusoidal function.  Here $t = 2/3$.  Thus, 
\begin{align*}
\frac{2}{3}b & = 2\pi\\
b & = 2\pi \cdot \frac{3}{2}\\
b & = 3\pi
\end{align*}
Substituting for $a$, $b$, and $d$ yields $x = 3\sin(3\pi t) - 2$.  
We can check this by substituting $0$ and $0.5$ for $t$.  If $t = 0$, then 
$$x = 3\sin(0) - 2 = -2$$
If $t = 0.5$, then
$$x = 3\sin\left(\frac{3\pi}{2}\right) - 2 = 3(-1) - 2 = -3 - 2 = -5$$
Both values agree with the graph.
