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

Explanation:

Step1: Determine the sinusoidal - function parameters

The general form of a sinusoidal function is $y = A\sin(B(x - C))+D$. The amplitude $A=\frac{69 - 51}{2}=9$, the vertical shift $D=\frac{69 + 51}{2}=60$. The period of a daily - temperature function is $T = 24$ hours, and since $T=\frac{2\pi}{B}$, then $B=\frac{2\pi}{24}=\frac{\pi}{12}$. The average temperature $y = 60$ first occurs at $x = 10$ (10 AM), so the phase - shift $C = 10$. Thus, the temperature function is $y=9\sin\left(\frac{\pi}{12}(x - 10)\right)+60$.

Step2: Set up the equation to solve for $x$

We want to find $x$ when $y = 58$. So we set up the equation $58=9\sin\left(\frac{\pi}{12}(x - 10)\right)+60$. First, subtract 60 from both sides: $58−60 = 9\sin\left(\frac{\pi}{12}(x - 10)\right)$, which gives $- 2=9\sin\left(\frac{\pi}{12}(x - 10)\right)$. Then $\sin\left(\frac{\pi}{12}(x - 10)\right)=-\frac{2}{9}$.

Step3: Solve for $x$

We know that if $\sin\theta=-\frac{2}{9}$, then $\theta=\sin^{-1}\left(-\frac{2}{9}\right)+2k\pi$ or $\theta=\pi-\sin^{-1}\left(-\frac{2}{9}\right)+2k\pi,k\in\mathbb{Z}$. We want the first non - negative solution for $x$. Let $\theta=\frac{\pi}{12}(x - 10)$. $\frac{\pi}{12}(x - 10)=\sin^{-1}\left(-\frac{2}{9}\right)$. $x - 10=\frac{12}{\pi}\sin^{-1}\left(-\frac{2}{9}\right)$. $\sin^{-1}\left(-\frac{2}{9}\right)\approx - 0.224$. $x - 10=\frac{12}{\pi}\times(-0.224)\approx - 0.85$. $x=10 - 0.85 = 9.15$.

Answer:

$9.15$