Question

modeling real - world scenarios with sine and cosine functions
temperature changes are observed over the course of three summer months and modeled by the function (f(x)=3cos(\frac{pi}{5}x+\frac{pi}{5}) + 25), where (x) represents the days after june 1 and (f(x)) represents the temperature in degrees celsius on day (x).
the first occurrence of the highest temperature in the cycle is day 9 with a temperature of 28°c. the first occurrence of the lowest temperature in the cycle is day with a temperature of 22°c.

Explanation:

Step1: Recall properties of cosine function

The general form of a cosine - function is $y = A\cos(Bx - C)+D$. For the function $f(x)=3\cos(\frac{\pi}{5}x+\frac{\pi}{5}) + 25$, where $A = 3$, $B=\frac{\pi}{5}$, $C=-\frac{\pi}{5}$, and $D = 25$. The maximum value of $\cos t$ is 1 and the minimum value is - 1.

Step2: Find the maximum value of $f(x)$

When $\cos(\frac{\pi}{5}x+\frac{\pi}{5})=1$, $f(x)$ reaches its maximum.
\[

$$\begin{align*} f(x)&=3\times1 + 25\\ &=28 \end{align*}$$

\]
We set $\frac{\pi}{5}x+\frac{\pi}{5}=2k\pi$, $k\in\mathbb{Z}$. Solving for $x$:
\[

$$\begin{align*} \frac{\pi}{5}x+\frac{\pi}{5}&=2k\pi\\ \frac{\pi}{5}(x + 1)&=2k\pi\\ x+1&=10k\\ x&=10k - 1 \end{align*}$$

\]
The first - occurrence ($k = 1$) gives $x = 9$.

Step3: Find the minimum value of $f(x)$

When $\cos(\frac{\pi}{5}x+\frac{\pi}{5})=-1$, $f(x)$ reaches its minimum.
\[

$$\begin{align*} f(x)&=3\times(-1)+25\\ &=22 \end{align*}$$

\]
We set $\frac{\pi}{5}x+\frac{\pi}{5}=(2k + 1)\pi$, $k\in\mathbb{Z}$. Solving for $x$:
\[

$$\begin{align*} \frac{\pi}{5}x+\frac{\pi}{5}&=(2k + 1)\pi\\ \frac{\pi}{5}(x + 1)&=(2k + 1)\pi\\ x+1&=5(2k + 1)\\ x&=10k+4 \end{align*}$$

\]
The first - occurrence ($k = 0$) gives $x = 4$.

Answer:

The first occurrence of the lowest temperature in the cycle is day 4 with a temperature of $22^{\circ}C$.