# Converting Polar to Cartesian Equations

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• Jun 10th 2008, 09:09 AM
farso
Converting Polar to Cartesian Equations
Hi there!

Im a little stuck on one of my questions and was hoping for a little bit of help.. So first, the question

Convert from polar to cartesian form, using double angle formula:

$r=\sqrt{1+cos2\theta}$

Ok, so things that I know:

$r^2=\sqrt{x^2+y^2}$
$x = r cos\theta$
$y = r sin\theta$

The double angle formula which is reffered to, i believe is
$cos2\theta = cos^2 \theta sin^2 \theta$
But i could be wrong

The furthest i have got to in the problem so far is

$r^2 = 1+cos^2 \theta sin^2 \theta$

And now im greatful if anyone is able to help me out please?

Thanks in advance!
• Jun 10th 2008, 09:19 AM
Moo
Hello,

\begin{aligned} \cos 2x &=\cos^2x-\sin^2x \\
&=1-2\sin^2x \\
&=2\cos^2x-1 \end{aligned}

this is the double angle formula for cos :)
• Jun 10th 2008, 09:56 AM
farso
Ah yeah, thanks.. I knew that but forgot the - sign in my original post.

I thought that the cos identity i chose was most appropriate as i could then relate x and y once i had moved things around further.

Still not sure on how things are to be done?
• Jun 10th 2008, 11:16 AM
Reckoner
Quote:

Originally Posted by farso
Still not sure on how things are to be done?

$r = \sqrt{1 + \cos2\theta}$

$\Rightarrow r = \sqrt{1 + \left(2\cos^2\theta - 1\right)}$

$\Rightarrow r = \sqrt{2\cos^2\theta} = \sqrt2\sqrt{\cos^2\theta} = \sqrt2\left\lvert\cos\theta\right\rvert$

$\Rightarrow r^2 = r\sqrt2\,\lvert\cos\theta\rvert\qquad\qquad\color{ red}\text{Multiplying both sides by }r$

$\Rightarrow r^2 = \sqrt2\,\lvert r\cos\theta\rvert\qquad\qquad\color{red}\text{Sinc e }r\text{ is nonnegative, }r\,\lvert\cos\theta\rvert = \lvert r\rvert\lvert\cos\theta\rvert = \lvert r\cos\theta\rvert$

$\Rightarrow x^2 + y^2 = \sqrt2\,\lvert x\rvert\qquad\qquad\color{red}\text{Substitution}$

$\Rightarrow y^2 = \sqrt2\,\lvert x\rvert - x^2$
• Jun 10th 2008, 03:05 PM
farso
Quote:

Originally Posted by Reckoner
$r = \sqrt{1 + \cos2\theta}$

$\Rightarrow r = \sqrt{1 + \left(2\cos^2\theta - 1\right)}$

$\Rightarrow r = \sqrt{2\cos^2\theta} = \sqrt2\sqrt{\cos^2\theta} = \sqrt2\left\lvert\cos\theta\right\rvert$

$\Rightarrow r^2 = r\sqrt2|\cos\theta|\qquad\qquad\color{red}\text{Mu ltiplying both sides by }r$

$\Rightarrow r^2 = \sqrt2|r\cos\theta|\qquad\qquad\color{red}\text{Si nce }r\text{ is nonnegative, }r|\cos\theta| = |r||\cos\theta| = |r\cos\theta|$

$\Rightarrow x^2 + y^2 = \sqrt2|x|\qquad\qquad\color{red}\text{Substitution }$

$\Rightarrow y^2 = \sqrt2|x| - x^2$

That seems like a really good explanation, however, I thought $r^2 =\sqrt{x^2+y^2}$ Am I not correct in this assumption?
• Jun 10th 2008, 05:05 PM
topsquark
Quote:

Originally Posted by farso
That seems like a really good explanation, however, I thought $r^2 =\sqrt{x^2+y^2}$ Am I not correct in this assumption?

No it is $r^2 = x^2 + y^2$. This is easy to remember since it is merely the Pythagorean Theorem. r is the hypotenuse and x and y are the legs.

-Dan
• Jun 10th 2008, 08:13 PM
Reckoner
Quote:

Originally Posted by farso
That seems like a really good explanation, however, I thought $r^2 =\sqrt{x^2+y^2}$ Am I not correct in this assumption?

As topsquark explained, you may use the Pythagorean theorem to get $r^2 = x^2 + y^2$ (you can also use this triangle to see where $\tan\theta = \frac yx$ comes from):