Chapter 9.4, Problem 8CFU

To determine

To write: The given equation in rectangular form

Answer to Problem 8CFU

The rectangular form of $3=r\cos \left(\theta -{60}^{\circ }\right)$ is $x+\sqrt{3}y-6=0$

Explanation of Solution

Given information:

The equation is $r=2\sec \left(\theta +\frac{\pi }{4}\right)$

Formula used:

  $\cos \left(A+B\right)=\cos A\cos B-\sin A\sin B$

Calculation:

The rectangular form of given equation can be obtained as,

  $\begin{array}{l}r=2\sec \left(\theta +\frac{\pi }{4}\right)\\ \frac{r}{\sec \left(\theta +\frac{\pi }{4}\right)}=2\\ r\cos \left(\theta +\frac{\pi }{4}\right)=2\\ r\left(\cos \theta \cos \frac{\pi }{4}-\sin \theta \sin \frac{\pi }{4}\right)=2\left[\text{difference identity for cosine}\right]\\ r\left(\cos \theta \cdot \frac{1}{\sqrt{2}}-\sin \theta \cdot \frac{1}{\sqrt{2}}\right)=2\left[\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}},\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}\right]\\ \frac{1}{\sqrt{2}}r\cos \theta -\frac{1}{\sqrt{2}}r\sin \theta =2\left[\text{distributive property}\right]\\ \frac{1}{\sqrt{2}}x-\frac{1}{\sqrt{2}}y=2\left[\text{polar to rectangular conversion}\right]\end{array}$

Therefore, the rectangular form of $r=2\sec \left(\theta +\frac{\pi }{4}\right)$ is $\frac{1}{\sqrt{2}}x-\frac{1}{\sqrt{2}}y=2$