Question

Use a graph to estimate the equations of all the vertical asymptotes of the curve

\(y = {\bf{tan}}\left( {{\bf{2sin}}\,x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \pi \le x \le \pi \)

Then find the exact equations of these asymptotes.

Short answer

The equations of asymptotes are \(x = \pm 0.90\), \(x = \pm 2.24\), \(x = \pm {\sin ^{ - 1}}\left( {\frac{\pi }{4}} \right)\), and \(x = \pm \left( {\pi - {{\sin }^{ - 1}}\frac{\pi }{4}} \right)\).

Step-by-step solution

Graph the function \(y = {\bf{tan}}\left( {{\bf{2sin}}\,x} \right)\)

Use the following steps to plot the graph of the given functions:

1. In the graphing calculator, select “STAT PLOT” and enter the equations \(\tan \left( {2\sin X} \right)\) in the \({Y_1}\) tab.
2. Enter the graph button in the graphing calculator.

The figure below represents the graph of the function\(y = \tan \left( {2\sin x} \right)\).

Find the asymptotes to the graph of \(y = {\bf{tan}}\left( {{\bf{2sin}}\,x} \right)\)

From the graph,the vertical asymptotes are \(x \approx \pm 0.90\) and \(x \approx \pm 2.24\).

The graph of the function has vertical asymptotes at; \(x = \frac{\pi }{2} + \pi n\).

Then, it can be represented as shown below:

\(\begin{array}{c}2\sin x &=& \frac{\pi }{2} + \pi n\\\sin x &=& \frac{\pi }{4} + \frac{\pi }{2}n\end{array}\)

As \( - 1 \le \sin x \le 1\), therefore the value is shown below:

\(\begin{array}{c}\sin x &=& \pm \frac{\pi }{4}\\x &=& \pm {\sin ^{ - 1}}\frac{\pi }{4}\end{array}\)

So, \(\pi - {\sin ^{ - 1}}\frac{\pi }{4}\) is the reference angle for \({\sin ^{ - 1}}\frac{\pi }{4}\). So, \(x = \pm \left( {\pi - {{\sin }^{ - 1}}\frac{\pi }{4}} \right)\) are the equation of vertical asymptotes.

Thus, the equations of asymptotes are \(x = \pm 0.90\), \(x = \pm 2.24\), \(x = \pm {\sin ^{ - 1}}\left( {\frac{\pi }{4}} \right)\), and \(x = \pm \left( {\pi - {{\sin }^{ - 1}}\frac{\pi }{4}} \right)\).