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

find an equation for a sinusoidal function that has period $pi$, amplitude $\frac{1}{2}$, and contains the point $(\frac{pi}{2},0)$. write your answer in the form $f(x)=asin(bx + c)+d$, where $a$, $b$, $c$, and $d$ are real numbers. $f(x)=$

find an equation for a sinusoidal function that has period $pi$, amplitude $\frac{1}{2}$, and contains the point $(\frac{pi}{2},0)$. write your answer in the form $f(x)=asin(bx + c)+d$, where $a$, $b$, $c$, and $d$ are real numbers. $f(x)=$

Answer

Answer:

$f(x)=\frac{1}{2}\sin(2x - \pi)$

Explanation:

Step1: Find the value of B

The period formula for $y = A\sin(Bx + C)+D$ is $T=\frac{2\pi}{|B|}$. Given $T = \pi$, we have $\pi=\frac{2\pi}{|B|}$. Solving for $B$, we get $|B| = 2$. Since we usually consider the positive - value for the basic form, $B = 2$.

Step2: Determine the value of A

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

Step3: Find the value of C and D

The general form is $f(x)=A\sin(Bx + C)+D$. Substitute $A=\frac{1}{2}$, $B = 2$, $x=\frac{\pi}{2}$, and $y = 0$ into $y=\frac{1}{2}\sin(2x + C)+D$. Since there is no vertical shift information, we assume $D = 0$. Then $0=\frac{1}{2}\sin(2\times\frac{\pi}{2}+C)$. So $0=\frac{1}{2}\sin(\pi + C)$. Using the trigonometric identity $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$, $\sin(\pi + C)=-\sin C$. Then $0 =-\frac{1}{2}\sin C$, which means $\sin C = 0$. We can choose $C=-\pi$ to satisfy the conditions. So the function is $f(x)=\frac{1}{2}\sin(2x - \pi)$.