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

Explanation:

Step1: Find the value of B

The period formula for a sinusoidal function $y = A\cos(Bx + C)+D$ is $T=\frac{2\pi}{|B|}$. Given $T = \pi$, we have $\pi=\frac{2\pi}{|B|}$. Solving for $B$ gives $|B| = 2$. Let's assume $B = 2$ for simplicity.

Step2: Determine the value of A

The amplitude of the function $y=A\cos(Bx + C)+D$ is $|A|$. Given that the amplitude is $\frac{1}{2}$, so $A=\pm\frac{1}{2}$.

Step3: Find the value of D

The mid - line of the cosine function $y = A\cos(Bx + C)+D$ is $y = D$. Since the cosine function oscillates between $D + A$ and $D - A$, and we have no information about vertical shift other than the point on the function. Let's substitute the known values into the function $y=A\cos(Bx + C)+D$. Substitute $x =-\frac{3\pi}{4}$, $y=-5$, $A =-\frac{1}{2}$ (we can choose the sign of $A$ based on the point's position, and we find that $A =-\frac{1}{2}$ works better in this case), and $B = 2$ into $y=A\cos(Bx + C)+D$. [ \begin{align*} -5&=-\frac{1}{2}\cos\left(2\times\left(-\frac{3\pi}{4}\right)+C\right)+D\ -5&=-\frac{1}{2}\cos\left(-\frac{3\pi}{2}+C\right)+D \end{align*} ] We know that $\cos\left(-\frac{3\pi}{2}+C\right)=\sin C$. Let $C = 0$ (we can choose the phase - shift $C$ to simplify the calculation as long as the function passes through the given point). Then [ \begin{align*} -5&=-\frac{1}{2}\sin(0)+D\ D&=- 5 \end{align*} ]

Answer:

$f(x)=-\frac{1}{2}\cos(2x)-5$