# Math Help - polar to cartesian

1. ## polar to cartesian

im having problems figuring out how my teacher came up with this solution.

the problem is to convert the following polar to cartesian
r=2-cos(2theta)

we are in agreement for the following steps
r=2-[(cos^2 (theta))-(sin^2(theta))] using the double angle identity

then he has
√(x^2 + y^2) = 2 - [(x/r)^2] + [(y/r)^2]
where did the (x/r) and (y/r) come from and why are they squared??

he goes on to further solve the problem as follows:
√(x^2 + y^2) * (x^2 + y^2) = 2(x^2 + y^2) - (x^2) + (y^2)

final step
√(x^2 + y^2) * (x^2 + y^2) = (x^2) + 3(y^2)

this guy is know to make some mistake ... OFTEN. but i dont want to write this problem off. can someone explain to me what is happening here and a better way to arrive at his answer .. or in the least tell me if the solution is correct.

thanks. its exam review

2. Originally Posted by osolage
im having problems figuring out how my teacher came up with this solution.

the problem is to convert the following polar to cartesian
r=2-cos(2theta)

we are in agreement for the following steps
r=2-[(cos^2 (theta))-(sin^2(theta))] using the double angle identity

then he has
√(x^2 + y^2) = 2 - [(x/r)^2] + [(y/r)^2]
where did the (x/r) and (y/r) come from and why are they squared??

he goes on to further solve the problem as follows:
√(x^2 + y^2) * (x^2 + y^2) = 2(x^2 + y^2) - (x^2) + (y^2)

final step
√(x^2 + y^2) * (x^2 + y^2) = (x^2) + 3(y^2)

this guy is know to make some mistake ... OFTEN. but i dont want to write this problem off. can someone explain to me what is happening here and a better way to arrive at his answer .. or in the least tell me if the solution is correct.

thanks. its exam review
I don't really understand your question...

Are you asking to change this equation

$r = 2 - \cos{(2\theta)}$

into Cartesians?

If so, recall that $r = \sqrt{x^2 + y^2}$.

Also, remember that $\cos{(2\theta)} = \cos^2{\theta} - \sin^2{\theta}$

Now since $x = r\cos{\theta}$ that means that $\cos{\theta} = \frac{x}{r}$

$= \frac{x}{\sqrt{x^2 + y^2}}$.

Also, $y = r\sin{\theta}$, which means that $\sin{\theta} = \frac{y}{r}$

$= \frac{y}{\sqrt{x^2 + y^2}}$.

So the equation becomes

$r = 2 - \cos{(2\theta)}$

$r = 2 - (\cos^2{\theta} - \sin^2{\theta})$

$r = 2 - \cos^2{\theta} + \sin^2{\theta}$

$\sqrt{x^2 + y^2} = 2 - \left(\frac{x}{\sqrt{x^2 + y^2}}\right)^2 + \left(\frac{y}{\sqrt{x^2 + y^2}}\right)^2$

$\sqrt{x^2 + y^2} = 2 - \frac{x^2}{x^2 + y^2} + \frac{y^2}{x^2 + y^2}$

$\sqrt{x^2 + y^2} = 2 + \frac{y^2 - x^2}{x^2 + y^2}$

$\sqrt{x^2 + y^2} = \frac{2x^2 + 2y^2 + y^2 - x^2}{x^2 + y^2}$

$\sqrt{x^2 + y^2} = \frac{x^2 + 3y^2}{x^2 + y^2}$...

You may wish to simplify it further...
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