6. a. sketch a graph of the function f given by f(θ)=3cos(θ + π). 4 pts. b. how does the graph of f compare…

6a.

Explanation:

Step1: Recall cosine - function properties

The general form of a cosine function is $y = A\cos(B\theta - C)+D$. For $f(\theta)=3\cos(\theta+\pi)$, $A = 3$, $B = 1$, $C=-\pi$, $D = 0$. The amplitude of $y = A\cos(B\theta - C)+D$ is $|A|$, so the amplitude of $f(\theta)$ is $|3| = 3$. The period of $y=\cos(B\theta)$ is $T=\frac{2\pi}{|B|}$, so the period of $f(\theta)$ is $\frac{2\pi}{|1|}=2\pi$. We know that $\cos(\theta+\pi)=-\cos(\theta)$. When $\theta = 0$, $f(0)=3\cos(\pi)=- 3$; when $\theta=\frac{\pi}{2}$, $f(\frac{\pi}{2})=3\cos(\frac{\pi}{2}+\pi)=0$; when $\theta=\pi$, $f(\pi)=3\cos(2\pi)=3$; when $\theta=\frac{3\pi}{2}$, $f(\frac{3\pi}{2})=3\cos(\frac{3\pi}{2}+\pi)=0$; when $\theta = 2\pi$, $f(2\pi)=3\cos(3\pi)=-3$. Then we can sketch the graph of $y = 3\cos(\theta+\pi)$ which is a cosine - wave with amplitude 3, period $2\pi$, and reflected about the $x$ - axis compared to $y = \cos(\theta)$.

6b.

Explanation:

Step1: Analyze the effect of the parameter 3

The parameter $A = 3$ in $y = A\cos(\theta)$ is the amplitude. For $h(\theta)=\cos(\theta)$ with amplitude 1 and $f(\theta)=3\cos(\theta+\pi)$, the graph of $f$ has an amplitude 3 times that of $h$. The values of $f(\theta)$ range from - 3 to 3, while the values of $h(\theta)$ range from - 1 to 1.

Step2: Analyze the effect of the parameter $\pi$

The parameter $\pi$ in $f(\theta)=3\cos(\theta+\pi)$ is a phase - shift. The general form of a phase - shift for $y=\cos(\theta)$ is $y=\cos(\theta - C)$. Here $C =-\pi$, which means the graph of $y = 3\cos(\theta)$ is shifted $\pi$ units to the left. Also, $\cos(\theta+\pi)=-\cos(\theta)$, so the graph of $f(\theta)$ is a reflection of $y = 3\cos(\theta)$ (and also of $h(\theta)=\cos(\theta)$) about the $x$ - axis.

7a.

Explanation:

Step1: Recall the range of the cosine function

The function is $f(d)=5.3\cos(\frac{2\pi d}{365})+12.28$. The range of the cosine function $y = \cos(x)$ is $[-1,1]$.

Step2: Find the maximum value

When $\cos(\frac{2\pi d}{365}) = 1$, $f(d)_{max}=5.3\times1 + 12.28=17.58$. In the context of the hours of daylight, this represents the maximum number of hours of daylight in Glasgow after June 21, 2017.

Step3: Find the minimum value

When $\cos(\frac{2\pi d}{365})=-1$, $f(d)_{min}=5.3\times(-1)+12.28 = 6.98$. In the context of the hours of daylight, this represents the minimum number of hours of daylight in Glasgow after June 21, 2017.

Answer:

Maximum: 17.58 hours, represents the most hours of daylight; Minimum: 6.98 hours, represents the least hours of daylight.

7b.

Explanation:

Step1: Recall the period formula for a cosine function

The general form of a cosine function is $y = A\cos(Bx - C)+D$, and its period is $T=\frac{2\pi}{|B|}$. For the function $f(d)=5.3\cos(\frac{2\pi d}{365})+12.28$, $B=\frac{2\pi}{365}$.

Step2: Calculate the period

Using the formula $T=\frac{2\pi}{|B|}$, we substitute $B=\frac{2\pi}{365}$ into it: $T=\frac{2\pi}{\frac{2\pi}{365}}=365$ days. In the context of the hours of daylight, this means that the pattern of the number of hours of daylight in Glasgow repeats every 365 days (a year).

Answer:

Period: 365 days, represents the time - interval for the daylight - hour pattern to repeat.

7c.

Explanation:

Step1: Recall the mid - line formula for a cosine function

For a cosine function of the form $y = A\cos(Bx - C)+D$, the mid - line is given by $y = D$. For the function $f(d)=5.3\cos(\frac{2\pi d}{365})+12.28$, $D = 12.28$. In the context of the hours of daylight, the mid - line $y = 12.28$ represents the average number of hours of daylight in Glasgow over the course of a year.

Answer:

Mid - line: $y = 12.28$, represents the average number of hours of daylight per day over a year.