# Thread: Converting Polar to Cartesian Equations

1. ## 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:

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

Ok, so things that I know:

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

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

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

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

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

2. Hello,

\displaystyle \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

3. 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?

4. Originally Posted by farso
Still not sure on how things are to be done?
$\displaystyle r = \sqrt{1 + \cos2\theta}$

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

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

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

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

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

$\displaystyle \Rightarrow y^2 = \sqrt2\,\lvert x\rvert - x^2$

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

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

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

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

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

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

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

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

6. Originally Posted by farso
That seems like a really good explanation, however, I thought $\displaystyle r^2 =\sqrt{x^2+y^2}$ Am I not correct in this assumption?
No it is $\displaystyle r^2 = x^2 + y^2$. This is easy to remember since it is merely the Pythagorean Theorem. r is the hypotenuse and x and y are the legs.

-Dan

7. Originally Posted by farso
That seems like a really good explanation, however, I thought $\displaystyle r^2 =\sqrt{x^2+y^2}$ Am I not correct in this assumption?
As topsquark explained, you may use the Pythagorean theorem to get $\displaystyle r^2 = x^2 + y^2$ (you can also use this triangle to see where $\displaystyle \tan\theta = \frac yx$ comes from):