a sine function has an amplitude of 3, period of 6\\pi, horizontal shift of \\frac{3\\pi}{2}, and vertical…

Explanation:

Step1: Write the general form of sine - function

The general form of a sine - function is $y = A\sin(B(x - C))+D$, where $A$ is the amplitude, $B$ is related to the period by $T=\frac{2\pi}{B}$, $C$ is the horizontal shift, and $D$ is the vertical shift. Given $A = 3$, $T = 6\pi$, $C=\frac{3\pi}{2}$, $D=-1$. First, find $B$ using the period formula: Since $T=\frac{2\pi}{B}$ and $T = 6\pi$, then $6\pi=\frac{2\pi}{B}$, so $B=\frac{2\pi}{6\pi}=\frac{1}{3}$. The sine - function is $y = 3\sin\left(\frac{1}{3}(x-\frac{3\pi}{2})\right)-1$.

Step2: Substitute $x = 2\pi$ into the function

Substitute $x = 2\pi$ into $y = 3\sin\left(\frac{1}{3}(x-\frac{3\pi}{2})\right)-1$: First, calculate the argument of the sine - function: $\frac{1}{3}(x-\frac{3\pi}{2})=\frac{1}{3}(2\pi-\frac{3\pi}{2})=\frac{1}{3}\times\frac{4\pi - 3\pi}{2}=\frac{\pi}{6}$. Then, $y = 3\sin\left(\frac{\pi}{6}\right)-1$. Since $\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}$, then $y = 3\times\frac{1}{2}-1=\frac{3}{2}-1=\frac{1}{2}$.

Answer:

$\frac{1}{2}$