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$t = 35sinleft(\frac{pi}{6}(m + 3)\right)+55$

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: Use the co - function identity

We know that $\cos(A)=\sin(A + \frac{\pi}{2})$. Given $t = 35\cos(\frac{\pi}{6}(m + 3))+55$. Let $A=\frac{\pi}{6}(m + 3)$. Then $\cos(\frac{\pi}{6}(m + 3))=\sin(\frac{\pi}{6}(m + 3)+\frac{\pi}{2})$.

Step2: Simplify the argument of the sine function

$\frac{\pi}{6}(m + 3)+\frac{\pi}{2}=\frac{\pi m}{6}+\frac{\pi}{2}+\frac{\pi}{2}=\frac{\pi m}{6}+\pi$. And $\sin(x+\pi)=-\sin(x)$. So $\sin(\frac{\pi m}{6}+\pi)=-\sin(\frac{\pi m}{6})$. Then $t = 35\cos(\frac{\pi}{6}(m + 3))+55=- 35\sin(\frac{\pi}{6}m)+55$.

Answer:

$t=-35\sin(\frac{\pi}{6}m)+55$