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Take a simple polar equation like r = θ/2 that graphs out to:

enter image description here

But, how would I achieve a rotation of the light-grey plot in this image (roughly 135 degrees)? Is there a way to easily shift the plot?

enter image description here

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    $\begingroup$ $r=(\theta-\frac{3\pi}{4})/2$ $\endgroup$ Jun 6, 2012 at 16:35
  • $\begingroup$ Rotation about the origin is simple in polar coordinates, just like translation to the right (or up, or both) is simple in the usual rectangular coordinates. $\endgroup$ Jun 6, 2012 at 16:42

2 Answers 2

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Just put $\theta-135^\circ$ in place of $\theta$. Or if you're working in radians, then the equivalent in radians.

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  • $\begingroup$ Ah, I don't know why I didn't think to try that. And thanks for the helpful note about radians, I was (apparently) working in radians and would have been stumped by trying degrees. $\endgroup$
    – Nick
    Jun 6, 2012 at 23:26
  • $\begingroup$ commons.wikimedia.org/wiki/File:FatTrifolium.svg $\endgroup$
    – Adam
    Dec 6, 2018 at 18:00
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A way to think about this is is that you want to shift all $\theta$ to $\theta'=\theta +\delta$, where $\delta$ is the amount by which you want to rotate. This question has a significance if you want to rotate some equation which is a function of theta. In the case $r=\theta$ that becomes $r=\theta+\delta$.

Of course if our independent variable in our polar equation was a non-identity function of $\theta$ you might be able to use the angle-sum indentities to help you out:

$$ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta $$

In case an anyone is trying to programme this in a Cartesian setting like I was trying to do (for a music visualizer) where I wanted my spiral's rotation to be a function of time. $r = \theta(t)$. Normally where solving $r=\theta$ or $\sqrt{x^2+y^2}=tan(\frac{\sin(\theta)}{\cos(\theta)})=tan(\frac{y}{x})$ you can substitute as follows.

$$ \sqrt{x^2+y^2}= tan(\frac{\sin(\theta+t)}{\cos(\theta+t)}) = tan(\frac{\sin\theta \cos t+\cos \theta \sin t}{\cos \theta \cos t - \sin \theta \sin t}) = tan(\frac{y \cos t +x\sin t }{x\cos t - y \sin t}) $$ /

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