sketch graphs f and g given by\nf(θ)= sin θ\ng(θ)= sin θ + 2\nuse two different colors for f and g.

Answer:

To sketch the graph of $y = f(\theta)=\sin\theta$:

1. Recall the key - points of the sine function:
   • When $\theta = 0$, $y=\sin(0)=0$.
   • When $\theta=\frac{\pi}{2}$, $y = \sin(\frac{\pi}{2})=1$.
   • When $\theta=\pi$, $y=\sin(\pi)=0$.
   • When $\theta=\frac{3\pi}{2}$, $y=\sin(\frac{3\pi}{2})=-1$.
   • When $\theta = 2\pi$, $y=\sin(2\pi)=0$.
   • Plot these points $(0,0),(\frac{\pi}{2},1),(\pi,0),(\frac{3\pi}{2}, - 1),(2\pi,0)$ and connect them with a smooth curve. The range of $y = \sin\theta$ is $[-1,1]$. To sketch the graph of $y = g(\theta)=\sin\theta + 2$:
2. Use the transformation rule for functions. The graph of $y = f(\theta)+c$ (where $c$ is a constant) is a vertical shift of the graph of $y = f(\theta)$. Here $c = 2$, so the graph of $y=\sin\theta+2$ is the graph of $y = \sin\theta$ shifted upwards by 2 units.
   • When $\theta = 0$, $y=\sin(0)+2=2$.
   • When $\theta=\frac{\pi}{2}$, $y=\sin(\frac{\pi}{2})+2=3$.
   • When $\theta=\pi$, $y=\sin(\pi)+2=2$.
   • When $\theta=\frac{3\pi}{2}$, $y=\sin(\frac{3\pi}{2})+2=1$.
   • When $\theta = 2\pi$, $y=\sin(2\pi)+2=2$.
   • Plot these points $(0,2),(\frac{\pi}{2},3),(\pi,2),(\frac{3\pi}{2},1),(2\pi,2)$ and connect them with a smooth curve. The range of $y=\sin\theta + 2$ is $[1,3]$. Use two different colors to distinguish the two graphs as required.

Explanation:

Step1: Identify key - points of $y = \sin\theta$

Use standard sine values at $\theta=0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi$.

Step2: Plot points for $y = \sin\theta$

Plot $(0,0),(\frac{\pi}{2},1),(\pi,0),(\frac{3\pi}{2}, - 1),(2\pi,0)$ and connect.

Step3: Apply vertical - shift rule

For $y=\sin\theta + 2$, shift $y = \sin\theta$ up by 2 units.

Step4: Identify key - points of $y=\sin\theta + 2$

Add 2 to $y$ - values of key - points of $y = \sin\theta$.

Step5: Plot points for $y=\sin\theta + 2$

Plot $(0,2),(\frac{\pi}{2},3),(\pi,2),(\frac{3\pi}{2},1),(2\pi,2)$ and connect.