Chapter 8.4, Problem 11E

Solution

To find: The equation used to convert an equation in terms of $\left(x,y\right)$ to an equation in terms of $\left(u,v\right)$ when the angle of rotation is $\theta ={\cos }^{-1}\left(\frac{3}{5}\right)$ .

The liner equation is $x=\frac{3u}{5}-\frac{4v}{5}$ and $y=\frac{4u}{5}+\frac{3v}{5}$ .

Given Information:

The axis of rotation is $\theta ={\cos }^{-1}\left(\frac{3}{5}\right)$ .

Formula:

The property of inverse trigonometry is $\cos \left({\cos }^{-1}\theta \right)=\theta$ .

The equations for the axis rotation are,

  $\begin{array}{l}x=u\cos \theta -v\sin \theta \\ y=u\sin \theta +v\cos \theta \end{array}$

Calculation:

Take cosine function in both side of the equation, $\theta ={\cos }^{-1}\left(\frac{3}{5}\right)$ and solve using the property, $\cos \left({\cos }^{-1}\theta \right)=\theta$ .

  $\begin{array}{l}\cos \theta =\cos \left({\cos }^{-1}\left(\frac{3}{5}\right)\right)\\ \cos \theta =\frac{3}{5}\end{array}$

Apply the trigonometric identity, ${\sin }^2\theta +{\cos }^2\theta =1$ and solve for $\sin \theta$ .

  $\begin{array}{c}{\sin }^2\theta +{\left(\frac{3}{5}\right)}^2=1\\ {\sin }^2\theta =1-\frac{9}{25}\\ {\sin }^2\theta =\frac{16}{25}\\ \sin \theta =\frac{4}{5}\end{array}$

Substitute $\sin \theta =\frac{4}{5}$ and $\cos \theta =\frac{3}{5}$ in above equation to find the equation in terms of $x$ .

  $\begin{array}{c}x=u\cdot \frac{3}{5}-v\cdot \frac{4}{5}\\ =\frac{3u}{5}-\frac{4v}{5}\end{array}$

Similarly, substitute $\sin \theta =\frac{4}{5}$ and $\cos \theta =\frac{3}{5}$ in above equation to find the equation in terms of $y$ .

  $\begin{array}{c}y=u\cdot \frac{4}{5}+v\cdot \frac{3}{5}\\ =\frac{4u}{5}+\frac{3v}{5}\end{array}$

Therefore, the required equations are $x=\frac{3u}{5}-\frac{4v}{5}$ and $y=\frac{4u}{5}+\frac{3v}{5}$ .