find an equation for a sinusoidal function that has period $\frac{5pi}{3}$, amplitude 3, and contains the…

Answer:

$f(x)=3\cos\left(\frac{6}{5}x+\frac{2\pi}{5}\right)$

Explanation:

Step1: Find the value of B

The period formula for $y = A\cos(Bx + C)+D$ is $T=\frac{2\pi}{|B|}$. Given $T = \frac{5\pi}{3}$, we have $\frac{2\pi}{|B|}=\frac{5\pi}{3}$. Cross - multiply: $2\pi\times3 = 5\pi\times|B|$. Then $|B|=\frac{6}{5}$. We can take $B=\frac{6}{5}$ for simplicity.

Step2: Determine the value of A

The amplitude of the function $y = A\cos(Bx + C)+D$ is $|A|$. Given amplitude is 3, so $A = 3$.

Step3: Find the value of D

Since there is no vertical shift information given other than the general form, and we assume the mid - line is $y = 0$, so $D = 0$.

Step4: Calculate the value of C

Substitute $x=-\frac{\pi}{3}$, $y=-1$, $A = 3$, $B=\frac{6}{5}$ and $D = 0$ into $y=A\cos(Bx + C)+D$. We get $-1=3\cos\left(\frac{6}{5}\times\left(-\frac{\pi}{3}\right)+C\right)$. Then $-1 = 3\cos\left(-\frac{2\pi}{5}+C\right)$, so $\cos\left(-\frac{2\pi}{5}+C\right)=-\frac{1}{3}$. A common solution is when $-\frac{2\pi}{5}+C = 0$ (using the basic cosine properties for simplicity), and $C=\frac{2\pi}{5}$.