# 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? • $r=(\theta-\frac{3\pi}{4})/2$ – Arturo Magidin Jun 6 '12 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. – André Nicolas Jun 6 '12 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 '12 at 23:26
• commons.wikimedia.org/wiki/File:FatTrifolium.svg – Adam Dec 6 '18 at 18:00

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.

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