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A Rational Parameterization of the Unit Circle

November 6, 2009

We’re all familiar with the usual trigonometric parameterization of the unit circle: Each point on x^2 + y^2 = 1 is given by (\cos \theta, \sin \theta) for some real \theta. Less well-known is the parameterization of the unit circle by rational functions.

The line through the point (-1, 0) with slope m is given by y = m(x+1). This line intersects the unit circle in one other point and as we vary m we strike every point on the unit circle. Here’s an illustration for a few values of m:

What are the coordinates of this point? Well, the point (x,y) satisfies both x^2 + y^2 = 1 and y = m(x+1) so we have

x^2 + (m(x+1))^2 = 1

\implies x^2 + m^2(x^2+2x+1) = 1

\implies (1+m^2)x^2 + 2m^2x + (m^2-1) = 0

\displaystyle{ \implies x = \frac{-2m^2 \pm \sqrt{4m^4 - 4(1+m^2)(m^2-1)}}{2(1+m^2)} }

\displaystyle{ \implies x = \frac{-2m^2 \pm \sqrt{4}}{2(1+m^2)}  }

\displaystyle{ \implies x = -1,\ \  \frac{1-m^2}{1+m^2} }

x = -1 corresponds to the point (-1, 0). When x = \frac{1-m^2}{1+m^2} we have

\displaystyle{ y^2 = 1 - x^2 }

\displaystyle{ y^2 = 1 - \left( \frac{1-m^2}{1+m^2} \right)^2 }

\displaystyle{ y^2 = \frac{(1+m^2)^2 - (1-m^2)^2}{(1+m^2)^2} }

\displaystyle{ y^2 = \frac{(1+2m^2 + m^4) - (1-2m^2 + m^4)}{(1+m^2)^2} }

\displaystyle{ y^2 = \frac{4m^2}{(1+m^2)^2} }

\displaystyle{ \implies y = \pm\frac{2m}{1+m^2} }

and we want the \pm to be + so that positive slopes correspond to the upper half of the circle as we illustrated above.

Therefore every point on the unit circle (other than (-1, 0)) is of the form

\displaystyle{ \left(\frac{1-m^2}{1+m^2},\ \frac{2m}{1+m^2}\right) } for some m. As m \rightarrow \pm \infty we have \left(\frac{1-m^2}{1+m^2},\ \frac{2m}{1+m^2}\right) \rightarrow (-1, 0).

We now have two parameterizations of the unit circle. How are they connected? For every m there is a \theta such that

\displaystyle{ \cos \theta = \frac{1-m^2}{1+m^2} \text{ and } \sin \theta = \frac{2m}{1+m^2} }

which gives

\displaystyle{ \tan \theta = \frac{2m}{1-m^2} }.

This calls to mind the tangent addition formula \tan(A+B) = \frac{\tan A + \tan B}{1-\tan A\tan B}. This suggests m = \tan(\theta/2) and \theta = 2\tan^{-1}(m). The usual calculations show that this is correct: If we start with m = \tan^{-1}(\theta/2) we get the desired \cos \theta and \sin \theta.

3 Comments leave one →
  1. lucky permalink
    October 28, 2011 11:54 am

    too bad you didn’t post a polynomial parameterization… now that would have been impressive

    • December 2, 2011 12:42 pm

      There is no non-constant real polynomial parameterization. You can see this by considering the degree of x^2 + y^2.


  1. The Weierstrass Substitution « Leaves of Math

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