A) Draw the graph of the functions of y= (1)/(2)x and y=3 ((1)/(2)x+(pi)/(3)) on same axes for which 0 x 5pi.

Here's how to approach this problem:

a) Draw the graph of the functions of $y=\sin \frac{1}{2}x$ and $y=3\sin \left(\frac{1}{2}x+\frac{\pi}{3}\right)$ on same axes for which $0 \le x \le 5\pi$.

Step 1: Analyze the first function, $y = \sin\left(\frac{1}{2}x\right)$.
This is a sine wave with:

• Amplitude: $A = 1$
• Period: $T = \frac{2\pi}{B} = \frac{2\pi}{1/2} = 4\pi$
• Phase shift: None The graph oscillates between $-1$ and 1.

Key points for $y = \sin\left(\frac{1}{2}x\right)$ in the interval $0 \le x \le 5\pi$:

$$\begin{array}{|c|c|c|} \hline x & \frac{1}{2}x & y = \sin\left(\frac{1}{2}x\right) \\ \hline 0 & 0 & 0 \\ \pi & \frac{\pi}{2} & 1 \\ 2\pi & \pi & 0 \\ 3\pi & \frac{3\pi}{2} & -1 \\ 4\pi & 2\pi & 0 \\ 5\pi & \frac{5\pi}{2} & 1 \\ \hline \end{array}$$

Step 2: Analyze the second function, $y = 3\sin\left(\frac{1}{2}x + \frac{\pi}{3}\right)$.
This is a sine wave with:

• Amplitude: $A = 3$
• Period: $T = \frac{2\pi}{B} = \frac{2\pi}{1/2} = 4\pi$
• Phase shift: To find the phase shift, factor out the coefficient of $x$ from the argument: $\frac{1}{2}x + \frac{\pi}{3} = \frac{1}{2}\left(x + \frac{2\pi}{3}\right)$. The phase shift is $\frac{2\pi}{3}$ units to the left. The graph oscillates between $-3$ and 3.

Key points for $y = 3\sin\left(\frac{1}{2}x + \frac{\pi}{3}\right)$ in the interval $0 \le x \le 5\pi$:

\begin{array}{|c|c|c|c|} \hline x & \frac{1}{2}x + \frac{\pi}{3} & \sin\left(\frac{1}{2}x + \frac{\pi}{3}\right) & y = 3\sin\left(\frac{1}{2}x + \frac{\pi}{3}\right) \\ \hline