Chapter 8.4, Problem 14E

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 $\theta$ such that $\cot 2\theta =\frac{-7}{24}$ .

The linear equations are $x=\frac{3u}{5}-\frac{4v}{5}$ and $y=\frac{4u}{5}+\frac{3v}{5}$ .

Given Information:

The axis of rotation is $\theta$ such that $\cot 2\theta =\frac{-7}{24}$ .

Formula:

The equations for the axis rotation are,

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

Trigonometric identities are,

  $\cos \theta =\sqrt{\frac{1+\cos 2\theta }{2}}$ and $\sin \theta =\sqrt{\frac{1-\cos 2\theta }{2}}$

Calculation:

It is known that when $\cot 2\theta =\frac{-7}{24}$ , then $\cos 2\theta =-\frac{7}{25}$ .

Apply the trigonometric identity $\cos \theta =\sqrt{\frac{1+\cos 2\theta }{2}}$ and substitute $\cos 2\theta =-\frac{7}{25}$ into the identity and simplify.

  $\begin{array}{c}\cos \theta =\sqrt{\frac{1+\left(-\frac{7}{25}\right)}{2}}\\ =\sqrt{\frac{1-\frac{7}{25}}{2}}\\ =\sqrt{\frac{9}{25}}\\ =\frac{3}{5}\end{array}$

Similarly, apply the trigonometric identity $\sin \theta =\sqrt{\frac{1-\cos 2\theta }{2}}$ and substitute $\cos 2\theta =-\frac{7}{25}$ into the identity and simplify.

  $\begin{array}{c}\sin \theta =\sqrt{\frac{1-\left(-\frac{7}{25}\right)}{2}}\\ =\sqrt{\frac{1+\frac{7}{25}}{2}}\\ =\sqrt{\frac{16}{25}}\\ =\frac{4}{5}\end{array}$

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

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

Substitute $\sin \theta =\frac{\text{4}}{5}$ and $\cos \theta =\frac{\text{3}}{5}$ 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 $y=\frac{4u}{5}+\frac{3v}{5}$ .