find an equation for a sinusoidal function that has period $2\\pi$, amplitude 1, and contains the point…

Explanation:

Step1: Determine the value of A

The amplitude of the cosine - type sinusoidal function $y = A\cos(Bx + C)+D$ is given by $|A|$. Since the amplitude is 1, we can choose $A = 1$ or $A=-1$.

Step2: Determine the value of B

The period $T$ of a sinusoidal function $y = A\cos(Bx + C)+D$ is given by the formula $T=\frac{2\pi}{|B|}$. Given that $T = 2\pi$, we have $2\pi=\frac{2\pi}{|B|}$. Solving for $B$, we get $|B| = 1$. Let's choose $B = 1$ for simplicity.

Step3: Determine the value of D

The mid - line of the sinusoidal function $y=A\cos(Bx + C)+D$ is $y = D$. For a standard cosine function with no vertical shift, $D = 0$.

Step4: Determine the value of C

We know that the function $f(x)=A\cos(Bx + C)+D$ passes through the point $(\frac{3\pi}{2},0)$ with $A = 1$, $B = 1$, and $D = 0$. Substitute these values into the function: [0=\cos(\frac{3\pi}{2}+C)] We know that $\cos\theta=0$ when $\theta=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$. So, $\frac{3\pi}{2}+C=\frac{\pi}{2}+k\pi$. Solving for $C$: [C=\frac{\pi}{2}-\frac{3\pi}{2}+k\pi=- \pi + k\pi] Let $k = 1$, then $C = 0$.

Answer:

$f(x)=\cos(x)$