outside temperature over a day can be modeled as a sinusoidal function. suppose you know the temperature…

Explanation:

Step1: Find the average and amplitude

The average temperature $A=\frac{38 + 72}{2}=55$ degrees. The amplitude $B=\frac{72 - 38}{2}=17$ degrees. The period of a daily - temperature function is $T = 24$ hours, so $\omega=\frac{2\pi}{T}=\frac{2\pi}{24}=\frac{\pi}{12}$. Let the sinusoidal function be $y = A + B\sin(\omega(x - h))$. Since the temperature first occurs at 8 AM (8 hours after midnight), $h = 8$. So the function is $y=55 + 17\sin(\frac{\pi}{12}(x - 8))$.

Step2: Set up the equation

We want to find $x$ when $y = 47$. So we set up the equation $47=55 + 17\sin(\frac{\pi}{12}(x - 8))$. First, simplify the equation: [ \begin{align*} 47-55&=17\sin(\frac{\pi}{12}(x - 8))\

• 8&=17\sin(\frac{\pi}{12}(x - 8))\ \sin(\frac{\pi}{12}(x - 8))&=-\frac{8}{17}\approx - 0.4706 \end{align*} ]

Step3: Solve for $x$

We know that $\sin\theta=-0.4706$. The inverse - sine function $\theta=\sin^{-1}(-0.4706)\approx - 0.492$ (in radians). Since the sine function has a period of $2\pi$, we consider the appropriate solution in the range of the daily - temperature model. The general solution for $\sin\theta=-0.4706$ is $\theta=\pi+\sin^{-1}(0.4706)$ or $\theta = 2\pi-\sin^{-1}(0.4706)$. We take the first non - negative solution for our context. $\theta=\pi+\sin^{-1}(0.4706)\approx3.1416 + 0.492=3.6336$. Now we set $\frac{\pi}{12}(x - 8)=3.6336$. [ \begin{align*} x-8&=\frac{3.6336\times12}{\pi}\ x-8&=\frac{43.6032}{\pi}\approx13.88\ x&=8 + 13.88=21.88 \end{align*} ]

Answer:

$21.88$