Question

graph each function for $0 \leq x \leq 2\pi$.

1. $y = \sin \frac{1}{6}x$

Explanation:

Step1: Find the period

The general form of a sine function is \(y = A\sin(Bx - C)+D\). For \(y=\sin(\frac{1}{6}x)\), \(B = \frac{1}{6}\). The period of a sine function \(y=\sin(Bx)\) is \(T=\frac{2\pi}{|B|}\).
Substituting \(B=\frac{1}{6}\) into the period formula:
\(T=\frac{2\pi}{\frac{1}{6}}= 12\pi\). But we are graphing for \(0\leq x\leq2\pi\).

Step2: Create a table of values

Choose some \(x\) - values in the interval \([0,2\pi]\):
When \(x = 0\), \(y=\sin(\frac{1}{6}\times0)=\sin(0) = 0\).
When \(x=\pi\), \(y=\sin(\frac{1}{6}\pi)=\frac{1}{2}\).
When \(x = 2\pi\), \(y=\sin(\frac{1}{6}\times2\pi)=\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\).

Step3: Plot the points and draw the graph

Plot the points \((0,0)\), \((\pi,\frac{1}{2})\), \((2\pi,\frac{\sqrt{3}}{2})\) and connect them with a smooth curve that has the shape of a sine - wave. Since the coefficient of \(\sin\) is \(1\) (i.e., \(A = 1\)), the amplitude is \(|A|=1\). The function \(y=\sin(\frac{1}{6}x)\) is a horizontal stretch of the basic sine function \(y = \sin(x)\).

Answer:

Plot the points \((0,0)\), \((\pi,\frac{1}{2})\), \((2\pi,\frac{\sqrt{3}}{2})\) and draw a smooth sine - wave curve through them for \(0\leq x\leq2\pi\).