Chapter 4.4, Problem 93E

Solution

To identify: The B and C the graphs of the sine and cosine functions are waveforms like the figure below. By correctly labeling the coordinates of points A, B, and C, you will get the graph of the function given.

  $y=3\cos 2xandA=\left(-\frac{\pi }{4},0\right)$

  

The point of B and C $\left(0,3\right)$

  $\left(\frac{3\pi }{4},0\right)$ respectively

Given information

The given equation is $y=3\cos 2xandA=\left(-\frac{\pi }{4},0\right)$

  

Explanation:

Consider the given information

The graph of sinusoidal function of the from $y=a\sin bxory=a\cos bx$ have the following characteristics

  $\begin{array}{c}amplitude=\left|a\right|\\ period=\frac{2\pi }{\left|b\right|}\end{array}$

Applying this concept to the given function $y=3\cos 2x$

We have $a=3andb=2$

So

  $\begin{array}{c}period=\frac{2\pi }{\left|2\right|}\\ =\frac{2\pi }{2}\\ =\pi \end{array}$

The amplitude and period of the sinusoidal function are $3and\pi$ respectively.

Use the given coordinate of point A and the computed amplitude and period from step 1 to determine the coordinate of the point B.

Notice that point B lies at the maximum and at $\frac{1}{4}$ of a period hence the coordinate of point B is given by

  $\begin{array}{c}\left(x+\frac{p}{4},a\right)\\ \left(-\frac{\pi }{4}+\frac{\pi }{4},3\right)\\ \left(0,3\right)\end{array}$

Hence point B lies at $\left(0,3\right)$

Notice that the cycle ends at point C which lies at the x − axis $\left(y=0\right)$ Using this information and the computed period from step 1 we have

  $\begin{array}{c}\left(x+p,0\right)\\ \left(-\frac{\pi }{4}+4,0\right)\\ \left(\frac{3\pi }{4},0\right)\end{array}$

Hence point C lies at $\left(\frac{3\pi }{4},0\right)$

The point of B and C $\left(0,3\right)$ $\left(\frac{3\pi }{4},0\right)$ respectively.