We are asked to find a cosine function that models the height of a person's hand as they spin their arm around their shoulder. The person's shoulder is 2m high, their arm length is 1m, and they spin their arm once every 8 seconds. The hand starts at their side.

1. Problem Description

We are asked to find a cosine function that models the height of a person's hand as they spin their arm around their shoulder. The person's shoulder is 2m high, their arm length is 1m, and they spin their arm once every 8 seconds. The hand starts at their side.

2. Solution Steps

The general form of the cosine function is

$y = A\cos(B(x - C)) + D$, where:

$A$ is the amplitude,

$B$ is related to the period by $P = \frac{2\pi}{B}$,

$C$ is the horizontal shift,

$D$ is the vertical shift.

The height of the shoulder is 2m, and the arm length is 1m. Therefore, the height of the hand varies between $2-1 = 1$m and $2+1 = 3$m.

The amplitude $A$ is half the difference between the maximum and minimum height, which is $\frac{3-1}{2} = 1$.

Since the hand starts at the side, the height starts at the minimum value, i.e., 1m.

Therefore, we can use a negative cosine function to model the height.

So, we have $A=-1$.

The period is 8 seconds, so $P = 8 = \frac{2\pi}{B}$. Solving for $B$, we get $B = \frac{2\pi}{8} = \frac{\pi}{4}$.

The vertical shift $D$ is the average height, which is $\frac{3+1}{2} = 2$.

The horizontal shift $C$ is 0, as the hand starts at its side at $t=0$.

So the equation is $y = -1\cos(\frac{\pi}{4}t) + 2$, which simplifies to $y = -\cos(\frac{\pi}{4}t) + 2$.

3. Final Answer

$-\cos(\frac{\pi}{4}t)+2$