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

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 $d = - 2sinleft(pileft(t+\frac{1}{2}\right)\right)+5$. 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?\n$d=-2cos(pi t)+5$\n$d=-2cosleft(pileft(t + \frac{1}{2}\right)\right)+5$\n$d = 2cos(pi t)+5$\n$d = 2cosleft(pileft(t+\frac{1}{2}\right)\right)+5$

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 $d = - 2sinleft(pileft(t+\frac{1}{2}\right)\right)+5$. 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?\n$d=-2cos(pi t)+5$\n$d=-2cosleft(pileft(t + \frac{1}{2}\right)\right)+5$\n$d = 2cos(pi t)+5$\n$d = 2cosleft(pileft(t+\frac{1}{2}\right)\right)+5$

Answer

Explanation:

Step1: Recall trigonometric identity

$\sin(A +\frac{\pi}{2})=\cos(A)$. Given $d = - 2\sin(\pi(t+\frac{1}{2}))+5$, expand the argument: $\pi(t+\frac{1}{2})=\pi t+\frac{\pi}{2}$. So $d=-2\sin(\pi t+\frac{\pi}{2}) + 5$.

Step2: Apply identity

Using $\sin(A+\frac{\pi}{2})=\cos(A)$ with $A = \pi t$, we get $d=-2\cos(\pi t)+5$.

Answer:

$d=-2\cos(\pi t)+5$