To solve this question, we're going to graph the function y=cos(1/3x) on the interval -2π≤x≤2π. Here are the steps:
1. Begin by identifying the key properties of the original cosine function y=cos(x) which has the general shape of a wave, cycles between -1 and 1 on the y-axis, and has a period of 2π (meaning it repeats its pattern every 2π along the x-axis).
2. Notice that the function we're graphing, y=cos(1/3x), has a coefficient of 1/3 inside the cosine function, which affects the period of the graph. In general, if the function has the form y=cos(bx), the period is given by (2π)/|b|. So in our case, the new period is 2π/|1/3| = 6π. This is because the coefficient 1/3 compresses the graph horizontally by a factor of 1/3.
3. Next, generate the x-values for our graph. As specified in the question, we want to graph the function on the interval -2π≤x≤2π. However, since our function has a period of 6π, we'll extend the range to -3π and 3π to capture one full period of the graph. Divide the interval into equal parts to plot the key points. It can be helpful to mark the quarters of the period, so at -3π/2, 0, 3π/2, and 3π.
4. To find the corresponding y-values, simply plug each x-value into the function y=cos(1/3x). Remember that the cosine function oscillates between 1 and -1, so the y-values will also be between these numbers.
5. Plot the points you found in the previous step on an x-y plane. For example, at x=0, cos(0) = 1, so we have the point (0,1). Similarly, at x = ±3π, we have the points (-3π, 1) and (3π, 1), and at x = ±3π/2, we have cos( ±π/2 ) = 0, so the points are (-3π/2, 0) and (3π/2, 0).
6. Connect the points with a smooth, continuous curve to form the graph of y = cos(1/3x). The shape should still resemble a wave, but it is horizontally stretched compared to the standard cosine function.
7. Always label your axes appropriately and also indicates the period of the function on the graph.
By following these steps, you should have successfully plotted the graph of y=cos(1/3x).