Hello-- I have four equations I simply can't figure out:
r=cot(theta)csc(theta)
r=8sin(theta)
r=2cos(theta)sin(theta)
rsin(theta+pi/6)=2
I tried squaring each of these in hopes of finding something to work with, but I'm lost...please help!!
Hello-- I have four equations I simply can't figure out:
r=cot(theta)csc(theta)
r=8sin(theta)
r=2cos(theta)sin(theta)
rsin(theta+pi/6)=2
I tried squaring each of these in hopes of finding something to work with, but I'm lost...please help!!
Hello, riverjib!
Do you mean "polar to rectangular"?
We need the conversion formulas:
.$\displaystyle \begin{array}{ccc}x & = & r\cos\theta \\ y & = & r\sin\theta \\ r & = & \sqrt{x^2+y^2}\end{array}$
We have: .$\displaystyle r \;=\;\frac{\cos\theta}{\sin\theta} + \frac{1}{\sin\theta}\quad\Rightarrow\quad r\sin\theta \:=\:\cos\theta + 1$\displaystyle (1)\;\;r\:=\:\cot\theta\csc\theta$
$
Multiply by $\displaystyle r\!:\;\;r(r\sin\theta) \:=\:r\cos\theta + r$
Substitute: .$\displaystyle \sqrt{x^2+y^2}\cdot y \:=\:x + \sqrt{x^2+y^2} $
Then: .$\displaystyle \sqrt{x^2+y^2}\cdot y - \sqrt{x^2+y^2} \:=\:x\quad\Rightarrow\quad(y-1)\sqrt{x^2+y^2} \:=\:x $
Square both sides: .$\displaystyle \boxed{(y-1)^2(x^2+y^2) \:=\:x^2}$
Multiply by $\displaystyle r\!:\;\;r^2\:=\:8\cdot r\sin\theta$$\displaystyle (2)\;\;r\:=\:8\sin\theta)$
Substitute: .$\displaystyle \boxed{x^2+y^2\:=\:8y}$
$\displaystyle (3)\;\;r\:=\:2\cos\theta\sin\theta$
Multiply by $\displaystyle r^2\!:\;\;r^3 \:=\:2(r\cos\theta)(r\sin\theta)$
Substitute: .$\displaystyle (x^2+y^2)^{\frac{3}{2}} \:=\:2xy
$
Square both sides: .$\displaystyle \boxed{(x^2+y^2)^3\:=\:4x^2y^2}$
We have: .$\displaystyle r\left(\sin\theta\cos\frac{\pi}{6} + \sin\frac{\pi}{6}\cos\theta\right) \:=\:2 \quad\Rightarrow\quad r\left(\frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta\right) \:=\:2$$\displaystyle (4)\;\;r\sin\left(\theta + \frac{\pi}{6}\right)\:=\:2$
Multiply by 2: .$\displaystyle r\left(\sqrt{3}\sin\theta + \cos\theta\right) \:=\:4\quad\Rightarrow\quad \sqrt{3}(r\sin\theta) + r\cos\theta \:=\:4$
Substitute: .$\displaystyle \boxed{\sqrt{3}\,y + x \:=\:4}$
Thank you! I did 40 problems like this successfully, and got stumped on these four almost two hours ago. I actually figured out just before you posted that I could multiply one side by r as long as I did the same to the other side (big duh! on my part), and that solved the problem of the missing r's. But I was still stuck on the fourth equation, and you helped me just as I was beginning to lose my mind.