find an equation for a sinusoidal function that has period $\frac{pi}{2}$, amplitude 2, and contains the…

Explanation:

Step1: Find the value of B

The period formula for a sinusoidal function $y = A\sin(Bx + C)+D$ is $T=\frac{2\pi}{|B|}$. Given $T = \frac{\pi}{2}$, we have $\frac{\pi}{2}=\frac{2\pi}{|B|}$. Solving for $|B|$ gives $|B| = 4$. We can choose $B = 4$ (without loss of generality).

Step2: Determine the value of A

The amplitude is given by $|A|$. Since the amplitude is 2, we can choose $A = 2$ (again, without loss of generality).

Step3: Find the value of D

The mid - line of the sinusoidal function is $y = D$. For a standard sinusoidal function, the range of $y=A\sin(Bx + C)+D$ is $[D - |A|,D + |A|]$. Since the amplitude $A = 2$, and we know that the maximum value of $\sin$ function is 1 and minimum is - 1. The mid - line value $D$ can be found considering the vertical shift. A sinusoidal function with amplitude 2 has a range of $[D - 2,D + 2]$. We know that the function passes through the point $(3\pi,4)$. Substituting into the general form $y=A\sin(Bx + C)+D$ with $A = 2$ and $B = 4$, we get $4=2\sin(4\times3\pi + C)+D$. Since $\sin(12\pi + C)=\sin(C)$ and the maximum value of $\sin$ is 1, when $\sin(4\times3\pi + C)=1$, we have $4=2\times1+D$, so $D = 2$.

Step4: Find the value of C

Substitute $x = 3\pi$, $y = 4$, $A = 2$, $B = 4$ and $D = 2$ into $y=A\sin(Bx + C)+D$. We get $4=2\sin(4\times3\pi + C)+2$. Simplify to $2\sin(12\pi + C)=2$, then $\sin(12\pi + C)=1$. Since $\sin(12\pi + C)=\sin(C)$ and $\sin\theta = 1$ when $\theta=\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}$, we can choose $C=\frac{\pi}{2}$.

Answer:

$f(x)=2\sin(4x+\frac{\pi}{2})+2$