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

Answer:

$\cos(2x+\pi)$

Explanation:

Step1: Determine the value of A

The amplitude is 1. For $y = A\cos(Bx + C)+D$, $|A|$ is the amplitude. So $A = 1$ or $A=- 1$.

Step2: Find the value of B

The period formula for $y=\cos(Bx + C)+D$ is $T=\frac{2\pi}{|B|}$. Given $T = \pi$, then $\pi=\frac{2\pi}{|B|}$. Solving for $B$, we get $|B| = 2$. Since we want a standard - form cosine function, we can take $B = 2$.

Step3: Determine the value of D

For a basic cosine - type function with no vertical shift, $D = 0$.

Step4: Find the value of C

The function is $y=\cos(2x + C)$. Substitute the point $x=-\frac{\pi}{2},y = - 1$ into the function: $-1=\cos(2\times(-\frac{\pi}{2})+C)$. So $-1=\cos(-\pi + C)$. We know that $\cos(-\pi + C)=-1$ when $-\pi + C=(2k + 1)\pi,k\in\mathbb{Z}$. Taking $k = 0$, we get $C=\pi$. So the function is $f(x)=\cos(2x+\pi)$.