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I know if $\sqrt{x^2+y^2} = x$, then the polar equation of this is $r=cos\theta$ So,how to get the rectangular form of this polar equation, is it complicate: $r=cos(10\theta)$

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2 Answers 2

You seem a little confused here. The actual change of variables, if you want to go from $(x,y)$ to $(r, \theta)$, is

$$ x = r \cos \theta, \quad y = r \sin \theta $$ and if you want to go from $(r, \theta)$ to $(x,y)$, you have to do $$ r = \sqrt{ x^2 + y^2 }, \quad \sin \theta = \frac y{\sqrt{x^2 + y^2}} \quad \text{ or } \quad \cos \theta = \frac x{\sqrt{x^2 + y^2}}. $$ from which you can derive $\theta$.

Hope that helps,

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up vote 0 down vote accepted

hint for my question, hope someone can conclude something from this and finish my question: $\cos n\theta = \sum_{k=0}^n \binom{n}{k} \cos^k \theta\,\sin^{n-k} \theta\,\cos\left(\frac{1}{2}(n-k)\pi\right)$

Maybe we could just set $cos^n\theta$ as $x^n$ anyone could simplifly that formula as much as possible?

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