the height, d, in feet of a ball suspended from a spring as a function of time, t, in seconds can be modeled…

the height, d, in feet of a ball suspended from a spring as a function of time, t, in seconds can be modeled by the equation d=-2sin(π(t + 1/2))+5. the ball is released from its lowest point at t = 0 seconds. using your knowledge of the general form of sine and cosine functions, which of the following equations can also model this situation? d=-2cos(πt)+5 d=-2cos(π(t + 1/2))+5 d=2cos(πt)+5 d=2cos(π(t + 1/2))+5

the height, d, in feet of a ball suspended from a spring as a function of time, t, in seconds can be modeled by the equation d=-2sin(π(t + 1/2))+5. the ball is released from its lowest point at t = 0 seconds. using your knowledge of the general form of sine and cosine functions, which of the following equations can also model this situation? d=-2cos(πt)+5 d=-2cos(π(t + 1/2))+5 d=2cos(πt)+5 d=2cos(π(t + 1/2))+5

Answer

Explanation:

Step1: Recall the co - function identity

We know that $\sin(A)=\cos(A - \frac{\pi}{2})$. Given the function $d=- 2\sin(\pi(t+\frac{1}{2})) + 5$.

Step2: Apply the co - function identity

Let $A=\pi(t+\frac{1}{2})$. Then $-2\sin(\pi(t+\frac{1}{2}))+5=-2\cos(\pi(t+\frac{1}{2})-\frac{\pi}{2})+5$.

Step3: Simplify the argument of the cosine function

Simplify $\pi(t+\frac{1}{2})-\frac{\pi}{2}=\pi t+\frac{\pi}{2}-\frac{\pi}{2}=\pi t$. So $-2\cos(\pi(t+\frac{1}{2})-\frac{\pi}{2})+5=-2\cos(\pi t)+5$.

Answer:

$d = - 2\cos(\pi t)+5$ (corresponding to the first option)