Addition Formulae For The Circle

The principal tool for the computation of addition formulae on the circle is Eulers formula \begin{equation} \label{eq:eulers} e^{i\theta} \space = \space \cos\theta \space + \space i\thinspace\sin\theta \end{equation} and its converse \begin{equation} \label{eq:circexp} \cos\theta \space = \space \tfrac 1 2 \left(e^{i\theta} + e^{-i\theta}\right),\qquad \sin\theta \space = \space \tfrac 1 {2i} \left(e^{i\theta} - e^{-i\theta}\right) \end{equation} As is very well known, the classic addition formulae for the circular functions can be obtained from the addition formula for the exponential function by, for example, equating the real and imaginery parts of both sides of the equation \begin{equation*} \cos(\theta_1 + \theta_2) \enspace + \enspace i \sin(\theta_1 + \theta_2) \space = \space e^{i(\theta_1 + \theta_2)} \space = \space e^{i\theta_1} e^{i\theta_2} \space = \space \left[\cos(\theta_1)\cos(\theta_2) - \sin(\theta_1)\sin(\theta_2)\right] \enspace + \enspace i\left[\cos(\theta_1)\sin(\theta_2) + \sin(\theta_1)\cos(\theta_2)\right] \end{equation*} This method easily extends to $n$ variables and can also be used to compute the symmetric addition formulae simply by putting $\theta_1 + \theta_2 + \ldots = 0$ on the left hand side. The computation is even simpler if written in terms of $x$ and $y$ coordinates on the unit circle. For reference, let $x_i = \cos(\theta_i)$ and $y_i = \sin(\theta_i)$ then the first four are \begin{equation*} \begin{aligned} n=1 : & \qquad x_1 \\ & \qquad y_1 \\ n=2 : & \qquad x_1 x_2 - y_1 y_2 \\ & \qquad x_1 y_2 + x_2 y_1 \\ n=3 : & \qquad x_1 x_2 x_3 - x_1 y_2 y_3 - x_2 y_1 y_3 - x_3 y_1 y_2 \\ & \qquad x_1 x_2 y_3 + x_1 x_3 y_2 + x_2 x_3 y_1 - y_1 y_2 y_3 \\ n=4 : & \qquad x_1 x_2 x_3 x_4 - x_1 x_2 y_3 y_4 - x_1 x_3 y_2 y_4 - x_1 x_4 y_2 y_3 - x_2 x_3 y_1 y_4 - x_2 x_4 y_1 y_3 - x_3 x_4 y_1 y_2 + y_1 y_2 y_3 y_4 \\ & \qquad x_1 x_2 x_3 y_4 + x_1 x_2 x_4 y_3 + x_1 x_3 x_4 y_2 + x_2 x_3 x_4 y_1 - x_1 y_2 y_3 y_4 - x_2 y_1 y_3 y_4 - x_3 y_1 y_2 y_4 - x_4 y_1 y_2 y_3 \end{aligned} \end{equation*}

However we will obtain the symmetric addition formulae for circular functions by a different method, similar to that previously used for cubic and quartic curves.

What is not so well known is that the circular functions have their own versions of the Frobenius-Stickelberger formulae^[1]. These formulae can be derived by writing a determinant of circular functions in exponential form using \eqref{eq:circexp}, then using elementary column operations to convert them to sums of Vandermonde determinants, then to products and finally back into circular functions.

\begin{equation} \label{eq:cfs2a} \begin{vmatrix} 1 & \cos(\theta_1) \\ 1 & \cos(\theta_2) \\ \end{vmatrix} \space = \space 2 \thinspace \sin\left(\frac{\theta_1 + \theta_2} 2\right)\sin\left(\frac{\theta_1 - \theta_2} 2\right) \end{equation}

\begin{equation*} LHS \space = \space \tfrac 1 2 \left( \begin{vmatrix} 1 & e^{i\theta_1} \\ 1 & e^{i\theta_2} \\ \end{vmatrix} \space + \space \begin{vmatrix} 1 & e^{-i\theta_1} \\ 1 & e^{-i\theta_2} \\ \end{vmatrix} \right) \space = \space - \tfrac 1 2 \left(1 - e^{-i\theta_1-i\theta_2}\right)\left(e^{i\theta_1} - e^{i\theta_2}\right) \space = \space RHS \end{equation*}

\begin{equation} \label{eq:cfs2b} \begin{vmatrix} 1 & \sin(\theta_1) \\ 1 & \sin(\theta_2) \\ \end{vmatrix} \space = \space -2 \thinspace \cos\left(\frac{\theta_1 + \theta_2} 2\right)\sin\left(\frac{\theta_1 - \theta_2} 2\right) \end{equation}

This equation is easily obtained by substituting $\theta_i = \theta_i - \tfrac 1 2 \pi$ in \eqref{eq:cfs2a} and then using the identity $\cos(\theta_i - \tfrac 1 2 \pi) = \sin(\theta_i)$

\begin{equation} \label{eq:cfs3} \begin{vmatrix} 1 & \cos(\theta_1) & \sin(\theta_1) \\ 1 & \cos(\theta_2) & \sin(\theta_2) \\ 1 & \cos(\theta_3) & \sin(\theta_3) \\ \end{vmatrix} \space = \space 4 \thinspace \sin\left(\frac{\theta_1 - \theta_2} 2\right)\sin\left(\frac{\theta_2 - \theta_3} 2\right)\sin\left(\frac{\theta_3 - \theta_1} 2\right) \end{equation}

\begin{equation*} LHS \space = \space - \frac 1 {2i} \begin{vmatrix} 1 & e^{i\theta_1} & e^{-i\theta_1} \\ 1 & e^{i\theta_2} & e^{-i\theta_2} \\ 1 & e^{i\theta_3} & e^{-i\theta_3} \\ \end{vmatrix} \space = \space - \frac {\left(e^{i\theta_1} - e^{i\theta_2}\right)\left(e^{i\theta_2} - e^{i\theta_3}\right)\left(e^{i\theta_3} - e^{i\theta_1}\right)} {2ie^{i(\theta_1+\theta_2+\theta_3)}} \space = \space RHS \end{equation*}

\begin{equation} \label{eq:cfs4a} \begin{vmatrix} 1 & \cos(\theta_1) & \sin(\theta_1) & \cos^2(\theta_1) \\ 1 & \cos(\theta_2) & \sin(\theta_2) & \cos^2(\theta_2) \\ 1 & \cos(\theta_3) & \sin(\theta_3) & \cos^2(\theta_3) \\ 1 & \cos(\theta_4) & \sin(\theta_4) & \cos^2(\theta_4) \\ \end{vmatrix} \space = \space 16 \thinspace \sin\left(\frac{\theta_1 + \theta_2 + \theta_3 + \theta_4} 2\right) \thinspace \prod_{i \lt j} \sin\left(\frac{\theta_i - \theta_j} 2\right) \end{equation}

\begin{equation*} LHS \space = \space -\frac 1 {8i} \left( \begin{vmatrix} 1 & e^{i\theta_1} & e^{-i\theta_1} & e^{2i\theta_1} \\ 1 & e^{i\theta_2} & e^{-i\theta_2} & e^{2i\theta_2} \\ 1 & e^{i\theta_3} & e^{-i\theta_3} & e^{2i\theta_3} \\ 1 & e^{i\theta_4} & e^{-i\theta_4} & e^{2i\theta_4} \\ \end{vmatrix} \space + \space \begin{vmatrix} 1 & e^{i\theta_1} & e^{-i\theta_1} & e^{-2i\theta_1} \\ 1 & e^{i\theta_2} & e^{-i\theta_2} & e^{-2i\theta_2} \\ 1 & e^{i\theta_3} & e^{-i\theta_3} & e^{-2i\theta_3} \\ 1 & e^{i\theta_4} & e^{-i\theta_4} & e^{-2i\theta_4} \\ \end{vmatrix} \right) \space = \space - \frac {\left[1 - e^{-i(\theta_1+\theta_2+\theta_3+\theta_4)}\right]\prod\limits_{j \lt k}\left(e^{i\theta_j} - e^{i\theta_k}\right)} {8ie^{i(\theta_1+\theta_2+\theta_3+\theta_4)}} \space = \space RHS \end{equation*}

\begin{equation} \label{eq:cfs4b} \begin{vmatrix} 1 & \cos(\theta_1) & \sin(\theta_1) & \cos(\theta_1)\sin(\theta_1) \\ 1 & \cos(\theta_2) & \sin(\theta_2) & \cos(\theta_2)\sin(\theta_2) \\ 1 & \cos(\theta_3) & \sin(\theta_3) & \cos(\theta_3)\sin(\theta_3) \\ 1 & \cos(\theta_4) & \sin(\theta_4) & \cos(\theta_4)\sin(\theta_4) \\ \end{vmatrix} \space = \space -16 \thinspace \cos\left(\frac{\theta_1 + \theta_2 + \theta_3 + \theta_4} 2\right) \thinspace \prod_{i \lt j} \sin\left(\frac{\theta_i - \theta_j} 2\right) \end{equation}

\begin{equation*} LHS \space = \space \frac 1 {8} \left( \begin{vmatrix} 1 & e^{i\theta_1} & e^{-i\theta_1} & e^{2i\theta_1} \\ 1 & e^{i\theta_2} & e^{-i\theta_2} & e^{2i\theta_2} \\ 1 & e^{i\theta_3} & e^{-i\theta_3} & e^{2i\theta_3} \\ 1 & e^{i\theta_4} & e^{-i\theta_4} & e^{2i\theta_4} \\ \end{vmatrix} \space - \space \begin{vmatrix} 1 & e^{i\theta_1} & e^{-i\theta_1} & e^{-2i\theta_1} \\ 1 & e^{i\theta_2} & e^{-i\theta_2} & e^{-2i\theta_2} \\ 1 & e^{i\theta_3} & e^{-i\theta_3} & e^{-2i\theta_3} \\ 1 & e^{i\theta_4} & e^{-i\theta_4} & e^{-2i\theta_4} \\ \end{vmatrix} \right) \space = \space \frac {\left[1 + e^{-i(\theta_1+\theta_2+\theta_3+\theta_4)}\right]\prod\limits_{j \lt k}\left(e^{i\theta_j} - e^{i\theta_k}\right)} {8e^{i(\theta_1+\theta_2+\theta_3+\theta_4)}} \space = \space RHS \end{equation*}

If in equation \eqref{eq:cfs4a} we put $x_i = \cos(\theta_i)$ and $y_i = \sin(\theta_i)$ for $i = 1,2,3,4$ and assume $\theta_1 + \theta_2 + \theta_3 + \theta_4 \equiv 0 \mod 2\pi$ then we have four points on a unit circle satisfying \begin{equation} \label{eq:cfs4x} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 \\ 1 & x_2 & y_2 & x_2^2 \\ 1 & x_3 & y_3 & x_3^2 \\ 1 & x_4 & y_4 & x_4^2 \\ \end{vmatrix} \space = \space 0 \end{equation} But this implies the four points also lie on an interpolating parabola with an implicit equation $A + Bx + Cy + Dx^2 = 0$ given by \begin{equation} \label{eq:parabola} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 \\ 1 & x_2 & y_2 & x_2^2 \\ 1 & x_3 & y_3 & x_3^2 \\ 1 & x & y & x^2 \\ \end{vmatrix} \space = \space 0 \end{equation} If we run a parabola, of the above form, through three rational points on a unit circle the fourth point of intersection will also be a rational point and the four points satisfy the symmetric addition formula \eqref{eq:cfs4x}. This is an analogue of the well known feature of cubic curves, where if we run a straight line through two rational points on a cubic curve, the third point of intersection will also be a rational point satisfying a symmetric addition formula.

More than that, by using an elementary column operation to add an arbitrary multiple of $\cos^2(\theta_i) + \sin^2(\theta_i) = 1$ to the fourth column of \eqref{eq:cfs4a} it can be seen that the interpolating parabola can be extended to any conic of the form $A + Bx + Cy + Sx^2 + Ty^2 = 0$ given by \begin{equation} \label{eq:conic} \begin{vmatrix} 1 & x_1 & y_1 & S x_1^2 \space + \space T y_1^2 \\ 1 & x_2 & y_2 & S x_2^2 \space + \space T y_2^2 \\ 1 & x_3 & y_3 & S x_3^2 \space + \space T y_3^2 \\ 1 & x & y & S x^2 \space + \space T y^2 \\ \end{vmatrix} \space = \space 0 \end{equation} Another way to define this conic is by requiring that it pass through some arbitrary distinct fifth point $x_5,y_5$, in which case it will be given by \begin{equation} \label{eq:conic2} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 & y_1^2 \\ 1 & x_2 & y_2 & x_2^2 & y_2^2 \\ 1 & x_3 & y_3 & x_3^2 & y_3^2 \\ 1 & x_5 & y_5 & x_5^2 & y_5^2 \\ 1 & x & y & x^2 & y^2 \\ \end{vmatrix} \space = \space 0 \end{equation} Conics of this form have their major and minor axis parallel to the $x$ or $y$ axis. So, if an axis-aligned conic section intersects a circle at four points, then the sum of the angles of those four points is congruent to zero. This is true whether it is an ellipse, parabola, hyperbola or even two straight lines.

I don't know if this observation has an official name, I like to call it the two parabola's theorem for the circle because it implies that if two parabola's drawn at right angles to one another intersect at four points then those four points lie on a circle.

Observe that the conic with angle $\alpha$ at infinity can be computed using formula \eqref{eq:conic2} by putting $(x_5,y_5) \space = \space (r\cos\alpha,r\sin\alpha)$ and letting $r \rightarrow \infty$, giving \begin{equation*} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 & y_1^2 \\ 1 & x_2 & y_2 & x_2^2 & y_2^2 \\ 1 & x_3 & y_3 & x_3^2 & y_3^2 \\ 0 & 0 & 0 & \cos^2\alpha & \sin^2\alpha \\ 1 & x & y & x^2 & y^2 \\ \end{vmatrix} \space = \space 0 \end{equation*} Further observe that this is equivalent to equation \eqref{eq:conic} with $S = -\sin^2\alpha$ and $T = \cos^2\alpha$ and is an hyperbola whose asymptotes have angle $\pm\alpha$.

In addition it can be seen that $S = \sin^2\alpha$ and $T = \cos^2\alpha$ yields an ellipse whose bounding rectangle has diagonals with angle $\pm\alpha$.

Putting $(x_5,y_5) \space = \space \lambda (x_1,y_1) \space + \space (1 - \lambda)(x_2,y_2)$ in \eqref{eq:conic2} and performing some simplifying row operations gives \begin{equation*} \begin{vmatrix} 1 & x_1 & y_1 & x_1^2 & y_1^2 \\ 1 & x_2 & y_2 & x_2^2 & y_2^2 \\ 1 & x_3 & y_3 & x_3^2 & y_3^2 \\ 0 & 0 & 0 & (x_1-x_2)^2 & (y_1-y_2)^2 \\ 1 & x & y & x^2 & y^2 \\ \end{vmatrix} \space = \space 0 \end{equation*} This implies that when $\alpha$ is equal to the angle of the line between $(x_1,y_1)$ and $(x_2,y_2)$ the hyperbola degenerates to a pair of straight lines, and likewise for other pairs of points.

Also note that if say, the first two points coincide, then matrix \eqref{eq:conic} is singular. It can be fixed by using the derivative of the first row with respect to $\theta_1$ for the second row, as follows \begin{equation*} \begin{vmatrix} 1 & x_1 & y_1 & S x_1^2 \space + \space T y_1^2 \\ 0 & -y_1 & x_1 & 2(T-S) x_1 y_1 \\ 1 & x_3 & y_3 & S x_3^2 \space + \space T y_3^2 \\ 1 & x & y & S x^2 \space + \space T y^2 \\ \end{vmatrix} \space = \space 0 \end{equation*} a process which can be justified by letting $\theta_2 \rightarrow \theta_1$ in \eqref{eq:cfs4a}. Similarly if other points coincide. Therefore to get nicely spaced field lines let $\alpha = \pi i/2N$ where $i = 1 \ldots N-1$ and choose

$N-1$ magenta hyperbola's given by $S=-\sin^2\alpha$ and $T=\cos^2\alpha$
$N-1$ blue ellipse's given by $S=\sin^2\alpha$ and $T=\cos^2\alpha$
2 gray parabola's given by $S=0,T=1$ and $S=1,T=0$, which is the limit case of both the hyperbola's and ellipses when $\alpha = 0, \pi/2$
3 pairs of pink straight lines given by the chords through pairs of points (tangents in the case of coincident points), which is the limit case for hyperbola's when $\alpha$ equals the angle of the chord / tangent

The ellipses and hyperbola's are graphed by rearranging $c_0 + c_1 x + c_2 y + c_3 x^2 + c_4 y^2 = 0$ into the form \begin{equation*} c_4\left(2 c_3 x \space + \space c_1\right)^2 \enspace + \enspace c_3\left(2 c_4 y \space + \space c_2 \right)^2 \enspace = \enspace c_1^2 c_4 \space + \space c_2^2 c_3 \space - \space 4 c_0 c_3 c_4 \end{equation*}

The identities \eqref{eq:cfs2a} and \eqref{eq:cfs4a} are just limit cases of previously computed formulae for order 2 elliptic functions $f$ with two simple poles. They exist for any $n$ and can be computed using the first $n$ terms of the sequence. \begin{equation*} 1, \space f, \space f', \space f^2, \space f f', \space f^3, \space f^2 f', \space f^4 \space \ldots \end{equation*} For the Jacobi $\cn$ function this basis becomes \begin{equation*} 1, \space \cn, \space \sn\dn, \space \cn^2, \space \cn\sn\dn, \space \cn^3, \space \cn^2\sn\dn, \space \cn^4 \space \ldots \end{equation*} which in the limit becomes this basis for the circular functions \begin{equation*} 1, \space \cos, \space \sin, \space \cos^2, \space \cos\sin, \space \cos^3, \space \cos^2\sin, \space \cos^4 \space \ldots \end{equation*} So for $n$ odd, $n = 2m + 1$ we have this formula \begin{equation} \label{eq:cfsodd} \begin{vmatrix} 1 & \cos\theta_1 & \sin\theta_1 & \ldots & \cos^m\theta_1\sin\theta_1 \\ 1 & \cos\theta_2 & \sin\theta_2 & \ldots & \cos^m\theta_2\sin\theta_2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & \cos\theta_n & \sin\theta_n & \ldots & \cos^m\theta_n\sin\theta_n \\ \end{vmatrix} \space = \space 2^{m(m+1)} \prod_{1 \le i \lt j \le n} \sin\left(\frac{\theta_i - \theta_j} 2\right) \end{equation} and for $n$ even, $n = 2m$ we have the formula \begin{equation} \label{eq:cfseven} \begin{vmatrix} 1 & \cos\theta_1 & \sin\theta_1 & \ldots & \cos^m\theta_1 \\ 1 & \cos\theta_2 & \sin\theta_2 & \ldots & \cos^m\theta_2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & \cos\theta_n & \sin\theta_n & \ldots & \cos^m\theta_n \\ \end{vmatrix} \space = \space 2^{m^2} \sin\left(\frac{\theta_1 + \theta_2 + \ldots + \theta_n} 2\right)\prod_{1 \le i \lt j \le n} \sin\left(\frac{\theta_i - \theta_j} 2\right) \end{equation}