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 = - 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 the co - function identity

We know that $\sin(A)=\cos(A - \frac{\pi}{2})$. Given the function $d=-2\sin(\pi(t+\frac{1}{2})) + 5$. First, expand the argument of the sine function: $\pi(t+\frac{1}{2})=\pi t+\frac{\pi}{2}$.

Step2: Apply the co - function identity

Using $\sin(A)=\cos(A - \frac{\pi}{2})$, we substitute $A = \pi t+\frac{\pi}{2}$. Then $\sin(\pi t+\frac{\pi}{2})=\cos((\pi t+\frac{\pi}{2})-\frac{\pi}{2})=\cos(\pi t)$. So, $-2\sin(\pi(t + \frac{1}{2}))+5=-2\cos(\pi t)+5$.

Answer:

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