# How to "rotate" a polar equation?

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

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?

• $r=(\theta-\frac{3\pi}{4})/2$ Jun 6, 2012 at 16:35
• 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. Jun 6, 2012 at 16:42

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

• 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.
– Nick
Jun 6, 2012 at 23:26
• commons.wikimedia.org/wiki/File:FatTrifolium.svg
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.
$$\sqrt{x^2+y^2}= tan(\frac{\sin(\theta+t)}{\cos(\theta+t)}) = tan(\frac{\sin\theta \cos t+\cos \theta \sin t}{\cos \theta \cos t - \sin \theta \sin t}) = tan(\frac{y \cos t +x\sin t }{x\cos t - y \sin t})$$ /