Chapter 1.4, Problem 38E

To determine

Determine the points on the graph.

Answer to Problem 38E

  $\boxed{y=\sin 2x}$ OR $\boxed{y=-\sin 2x}$

Explanation of Solution

Given information:

The first graph in the figure is that of $y=\sin 45x$ as displayed by a $TI-83$ graphing calculator. It is inaccurate and so, to help explain its appearance, we re plot the curve in dot mode in the second graph. What two sine curves does the calculator appear to be plotting? Show that each point on the graph of $y=\sin 45x$ that the $TI-83$ chooses to plot is in fact on one of these two curves. (The $TI-83$ ’s graphing window is $95$ pixels wide,)

  

Calculation:

The graphs are drawn in the interval $[a,b]=[0,2\pi ]$ .

The number of gaps between two pixels is $95-1=94$ .

The distance between pixels is given by the following relation:

  $\begin{array}{l}d=\frac{b-a}{number-of-gaps}\\ =\frac{2\pi -0}{94}\\ =\frac{2\pi }{94}\end{array}$

The $x$ values of the plots deawn by calulator are given by $x=a+dn$ .

  $x=a+dn$

  $\begin{array}{l}=0+\frac{2\pi }{94}.n\\ x=\frac{2\pi }{94}n\end{array}$

For the curve $y=\sin 45x$ , the actual points plotted by the calculator are as follows:

  $(x,y)=\left(\frac{2\pi }{94}.n,\sin \left(45.\frac{2\pi }{94}.n\right)\right)$

Rewrite $y=\sin 45x$ as follows:

  $\sin \left(45.\frac{2\pi }{94}.n\right)=\sin \left(\left(47-2\right).\frac{2\pi }{94}.n\right)$$\begin{array}{l}=\sin \left(\left(47\right).\frac{2\pi }{94}.n-\left(2\right).\frac{2\pi }{94}.n\right)\\ =\sin \left(\pi n-2.\frac{2\pi }{94}.n\right)\end{array}$

Now use the formula $\sin (A-B)=\sin A\cos B-\cos A\sin B$

  $\begin{array}{l}\sin \left(\pi n-2.\frac{2\pi }{94}.n\right)=\sin \left(\pi n\right).\cos \left(2.\frac{2\pi }{94}.n\right)-\cos \left(\pi n\right).\sin \left(2.\frac{2\pi }{94}.n\right)\\ =0\left(\cos \left(2.\frac{2\pi }{94}.n\right)\right)-[-{\left(1\right)}^n].\sin \left(2.\frac{2\pi }{94}.n\right)\sin ce\{\begin{array}{l}\sin (\pi n)=0\\ \cos (\pi n)={(}^{-}\end{array}\\ =-{\left(-1\right)}^n\sin \left(2.\frac{2\pi }{94}.n\right)\\ =-\left(\pm 1\right)\sin \left(2.\frac{2\pi }{94}.n\right)\\ =\pm \sin \left(2.\frac{2\pi }{94}.n\right)\end{array}$

Hence, from this it is observe that the $y$ values and the points plotted for $y=\sin 45x$ lie on either $y=\sin 2x$ or $y=-\sin 2x$ .