Answer:
C.
[tex]y=\frac{13}{2}sin2x[/tex]
Explanation:
We were given the following information:
[tex]\begin{gathered} amplitude=\frac{13}{2} \\ period=2\pi \\ phase\text{ }shift=0 \\ midline:y=0 \end{gathered}[/tex]
We will proceed to derive the sinusoidal equation for this as shown below:
[tex]\begin{gathered} \text{We have the base model to be:} \\ y=sinx \\ \text{Inputting the amplitude into the equation, we have:} \\ y=\frac{13}{2}sinx \\ \text{Fitting in the period, we have:} \\ For:k>0 \\ y=\frac{13}{2}sinkx \\ k=2 \\ \text{The equation becomes:} \\ y=\frac{13}{2}sin2x \\ \text{Since the phase shift is ''0'', the equation of this function is given by:} \\ y=\frac{13}{2}s\imaginaryI n2x \\ \\ \therefore y=\frac{13}{2}s\imaginaryI n2x \end{gathered}[/tex]
Hence, the correct option is C
