the average daily temperature, $t$, in degrees fahrenheit for a city as a function of the month of the year…
the average daily temperature, $t$, in degrees fahrenheit for a city as a function of the month of the year, $m$, can be modeled by the equation $t = 35cosleft(\frac{pi}{6}(m + 3)\right)+55$, where $m = 0$ represents january 1, $m = 1$ represents february 1, $m = 2$ represents march 1, and so on. which equation also models this situation?\n$t=-35sinleft(\frac{pi}{6}m\right)+55$\n$t=-35sinleft(\frac{pi}{6}(m + 3)\right)+55$\n$t = 35sinleft(\frac{pi}{6}m\right)+55$\n$t = 35sinleft(\frac{pi}{6}(m + 3)\right)+55$
Answer
Explanation:
Step1: Recall the co - function identity
We know that $\cos(A)=\sin\left(A + \frac{\pi}{2}\right)$. Given $t = 35\cos\left(\frac{\pi}{6}(m + 3)\right)+55$, let $A=\frac{\pi}{6}(m + 3)$. Then $\cos\left(\frac{\pi}{6}(m + 3)\right)=\sin\left(\frac{\pi}{6}(m + 3)+\frac{\pi}{2}\right)$.
Step2: Simplify the argument of the sine function
Simplify $\frac{\pi}{6}(m + 3)+\frac{\pi}{2}=\frac{\pi m}{6}+\frac{\pi}{2}+\frac{\pi}{2}=\frac{\pi m}{6}+\pi$. And $\sin(x+\pi)=-\sin(x)$. So $\sin\left(\frac{\pi}{6}(m + 3)+\frac{\pi}{2}\right)=-\sin\left(\frac{\pi}{6}m\right)$.
Step3: Substitute back into the original equation
Substituting $\cos\left(\frac{\pi}{6}(m + 3)\right)=-\sin\left(\frac{\pi}{6}m\right)$ into $t = 35\cos\left(\frac{\pi}{6}(m + 3)\right)+55$, we get $t=-35\sin\left(\frac{\pi}{6}m\right)+55$.
Answer:
$t=-35\sin\left(\frac{\pi}{6}m\right)+55$