Chapter 13.6, Problem 48PPSE

Solution

a.

Graph the function $y=\frac{1-\cos x}{\sin x}$ using the window $\left[-3\pi ,3\pi \right]$ by $\left[-4,4\right]$ .

The graph of the given function is drawn.

Given:

A function $y=\frac{1-\cos x}{\sin x}$ .

Calculation:

The graph of the function $y=\frac{1-\cos x}{\sin x}$ on a graphing calculator using the window $\left[-3\pi ,3\pi \right]$ by $\left[-4,4\right]$ is shown below:

  

It is a tangent curve of the form $y=a\tan bx$ .

Conclusion:

The graph of the function is drawn.

b.

The period of the curve and the value of a .

The period of the curve is $2\pi$ and the value of a is 1.

Given:

Graph of $y=\frac{1-\cos x}{\sin x}$ is the tangent curve of the form $y=a\tan bx$ .

Concept used:

The period of a function is known as the distance after which the graph of the function gets repeated.

For any function $y=a\tan bx$ , the period is $\frac{\pi }{b}$ .

Calculation:

As shown the graph in part (a), the period of the function $y=\frac{1-\cos x}{\sin x}$ is $2\pi$ after which the graph is getting repeated.

Since the period of function $y=a\tan bx$ is $\frac{\pi }{b}$ , then

  $\begin{array}{l}2\pi =\frac{\pi }{b}\\ b=\frac{\pi }{2\pi }\\ b=\frac{1}{2}\end{array}$

To calculate the value of a , take a point from the graph and substitute these coordinates in the function $y=a\tan bx$ .

Let us take the point $\left(\frac{\pi }{2},1\right)$ . On substituting these values in the function:

  $\begin{array}{c}y=a\tan bx\\ 1=a\tan \frac{1}{2}\left(\frac{\pi }{2}\right)\\ 1=a\tan \left(\frac{\pi }{4}\right)\\ a=1\text{ Since }\tan \left(\frac{\pi }{4}\right)=1\end{array}$

Conclusion:

The period of the curve is $2\pi$ and the value of a is 1.

c.

The x -coordinate halfway between a removable discontinuity and the right side asymptote and the corresponding y -coordinate.

The x -coordinate halfway between a removable discontinuity (at $x=0$ ) and the right side asymptote (at $x=\pi$ ) will be $x=\frac{\pi }{2}$ and the corresponding y -coordinate will be $y=1$ .

Given:

A function:

  $y=\frac{1-\cos x}{\sin x}$

Concept used:

A function has removable discontinuity at $x=a$ when $f\left(x\right)\ne f\left(a\right)$ .

An asymptote is a straight line which visits the curve but never intersects it. The asymptote moves toward infinity.

Calculation:

The graph of the function $y=\frac{1-\cos x}{\sin x}$ is shown below:

  

At $x=0$ , the function $y=\frac{1-\cos x}{\sin x}$ has removable discontinuity since $\sin \left(0\right)=0$ .

The asymptote of the function $y=\frac{1-\cos x}{\sin x}$ is at $x=-\pi$ and $x=\pi$ where it is not intersecting the curve and moving toward infinity.

The asymptote at the right side of removable discontinuity is $x=\pi$ .

Thus, the required x -coordinate halfway between a removable discontinuity (at $x=0$ ) and the right side asymptote (at $x=\pi$ ) will be $x=\frac{\pi }{2}$ and the corresponding y -coordinate will be $y=1$ .

Conclusion:

The required x -coordinate halfway between a removable discontinuity (at $x=0$ ) and the right side asymptote (at $x=\pi$ ) will be $x=\frac{\pi }{2}$ and the corresponding y -coordinate will be $y=1$ .

d.

An equivalent function of the form $y=a\tan bx$ .

The equivalent function of the form $y=a\tan bx$ is $y=\tan \left(\frac{x}{2}\right)$ .

Given:

A function:

  $y=a\tan bx$

Calculation:

From part (b), the value of a is 1 and the value of b is $\frac{1}{2}$ .

On substituting these values of a and b into the given function $y=a\tan bx$ ,

  $\begin{array}{c}y=a\tan bx\\ =1\cdot \tan \frac{1}{2}x\\ y=\tan \frac{x}{2}\end{array}$

Conclusion:

An equivalent function of the form $y=a\tan bx$ is $y=\tan \frac{x}{2}$ .