Changing polar equations to rectangular equations?

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To convert polar equations to rectangular equations, the relationships x = r cos(θ) and y = r sin(θ) are essential. The equation r = θ/2 can be transformed by multiplying both sides by r, leading to r² = r sin(θ), which simplifies to x² + y² = y, representing a circle. For r = 6cos(θ) + sin(θ), multiplying by r gives r² = 6r cos(θ) + r sin(θ), resulting in x² + y² = 6x + y, also a circle. Lastly, for r² sin(2θ) = 2, using sin(2θ) = 2sin(θ)cos(θ) aids in conversion. Understanding these transformations is crucial for solving polar equations effectively.
steener
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changing polar equations to rectangular equations?

Can somebody please explain to me, how I would convert:

?=?/2 into a rectangular equation?

Along with: r=sin?, r=6cos+sin?, r(squared)sin2?=2


Your help would be greatly appreciated!
 
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steener said:
Can somebody please explain to me, how I would convert:

?=?/2 into a rectangular equation?

Along with: r=sin?, r=6cos+sin?, r(squared)sin2?=2


Your help would be greatly appreciated!

Unfortunately your "special characters" just show up as "?" to me. I would guess that the ? in the last three are "theta": \theta in LaTex, but I have no idea what the "?" in ?= ?/2 are- I presume they are different or the equation is trivial.

I presume that you know (or else you wouldn't be attempting these problems) that x= r cos(\theta) and y= r sin(\theta). Looking at the first one, my thought would be to multiply both sides by r: r^2= r sin(\theta) which, since r^2= r^2(cos^2(\theta)+ sin^2(\theta)= x^2+ y^2, is just x^2+ y^2= y, the equation of a circle.

For r= 6cos(\theta)+ sin(theta), same thing: multiply both sides by r to get r^2= 6r cos(\theta)+ r sin(\theta)= x^2+ y^2= 6x+ y, again the equation of a circle.

For r^2 sin(2\theta)= 2, use the fact that sin(2\theta)= 2sin(\theta)cos(\theta).
 

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