KindiakMath

Defending Rational Trigonometry

Let \mathbb K be a field. In the case \mathbb K = \mathbb Q \subseteq \mathbb R \subseteq \mathbb C, we can define the usual inner product \langle \cdot, \cdot \rangle on \mathbb K^n via

\displaystyle \langle \mathbf u, \mathbf v \rangle = \sum_{i=1}^n u_i \bar{v}_i = \mathbf v^* \mathbf u = \mathbf v^* \mathbf I_n \mathbf u \in \mathbb K.

Furthermore, in the case \mathbb K = \mathbb Q \subseteq \mathbb R, \mathbf v^* \mathbf u = \mathbf u^* \mathbf v due to symmetry. In particular, setting \mathbf u = \mathbf v, we get

\displaystyle \langle \mathbf v, \mathbf v \rangle = \sum_{i=1}^n v_i \bar{v}_i = \sum_{i=1}^n |v_i|^2 \in \mathbb K.

It is the square that motivates us to call the map Q(\mathbf v) = \langle \mathbf v, \mathbf v \rangle a quadratic form, that satisfies the following properties:

Definition 1. Let V be any vector space over \mathbb K. Let Q : V \to \mathbb K be a map and define the map [\cdot, \cdot] : V \times V \to \mathbb K by

[\mathbf u,\mathbf v] := \frac 12 (Q(\mathbf u + \mathbf v) - Q(\mathbf u) - Q(\mathbf v)).

We call Q a quadratic form if

  • For any \alpha \in \mathbb K, Q(\alpha \mathbf v) = \alpha^2 Q(\mathbf v).
  • The map [ \cdot , \cdot ] is symmetric and bi-linear.

Lemma 1. For any matrix \mathbf A \in \mathcal M_{n \times n}(\mathbb K), the map Q_{\mathbf A} : \mathbb K^n \to \mathbb K defined by Q_{\mathbf A}(\mathbf v) = \mathbf v^{\mathrm T} \mathbf A \mathbf v is a quadratic form. Furthermore, there exists a unique symmetric matrix \mathbf B such that Q_{\mathbf A} = Q_{\mathbf B}.

Proof. For the quadratic form claim, we first note that

\begin{aligned} 2[\mathbf u, \mathbf v]_{\mathbf A} &= Q_{\mathbf A}(\mathbf u + \mathbf v) - Q_{\mathbf A}(\mathbf u) - Q_{\mathbf A}(\mathbf v) \\ &= (\mathbf u + \mathbf v)^{\mathrm T} \mathbf A (\mathbf u + \mathbf v) - \mathbf u^{\mathrm T} \mathbf A \mathbf u - \mathbf v^{\mathrm T} \mathbf A \mathbf v \\ &= (\mathbf u^{\mathrm T} + \mathbf v^{\mathrm T}) (\mathbf A \mathbf u + \mathbf A \mathbf v) - \mathbf u^{\mathrm T} \mathbf A \mathbf u - \mathbf v^{\mathrm T} \mathbf A \mathbf v \\ &= \mathbf u^{\mathrm T}\mathbf A \mathbf u + \mathbf u^{\mathrm T}\mathbf A \mathbf v + \mathbf v^{\mathrm T}\mathbf A \mathbf u + \mathbf v^{\mathrm T}\mathbf A \mathbf v - \mathbf u^{\mathrm T} \mathbf A \mathbf u - \mathbf v^{\mathrm T} \mathbf A \mathbf v \\ &= \mathbf u^{\mathrm T}\mathbf A \mathbf v + \mathbf v^{\mathrm T}\mathbf A \mathbf u \\ &= \mathbf u^{\mathrm T}\mathbf A \mathbf v + (\mathbf v^{\mathrm T}\mathbf A \mathbf u)^{\mathrm T} \\ &= \mathbf u^{\mathrm T}\mathbf A \mathbf v + \mathbf u^{\mathrm T} \mathbf A^{\mathrm T}\mathbf v \\ &= \mathbf u^{\mathrm T}(\mathbf A + \mathbf A^{\mathrm T})\mathbf v\end{aligned}

Therefore, by bookkeeping, [ \cdot , \cdot ]_{\mathbf A} is bilinear. For the symmetric matrix claim, we first note that for any \mathbf v,

\begin{aligned} Q_{\mathbf A}(\mathbf v) &= [\mathbf v, \mathbf v]_{\mathbf A} = \mathbf v^{\mathrm T} \left( \frac 12 (\mathbf A + \mathbf A^{\mathrm T}) \right) \mathbf v = \mathbf v^{\mathrm T} \mathbf B \mathbf v = Q_{\mathbf B}(\mathbf v), \end{aligned}

where \mathbf B := \frac 12 (\mathbf A + \mathbf A^{\mathrm T}) is a desired symmetric matrix. Furthermore,

[\mathbf e_i, \mathbf e_j]_{\mathbf A} = a_{ij} + a_{ji},

so that [\mathbf e_i, \mathbf e_j]_{\mathbf B} = 2b_{ij}. Finally, if there exists another symmetric matrix \mathbf C such that Q_{\mathbf A} = Q_{\mathbf C}, then

b_{ij} = \frac 12 [\mathbf e_i, \mathbf e_j]_{\mathbf B} = \frac 12 [\mathbf e_i, \mathbf e_j]_{\mathbf C} = c_{ij},

so that \mathbf B = \mathbf C, as required.

This means that any symmetric matrix defines a quadratic form. It turns out that the reverse happens to be true too: any quadratic form can be defined in terms of a symmetric matrix \mathbf A.

Lemma 3. For any quadratic form Q : \mathbb K^n \to \mathbb K, there exists a unique symmetric matrix \mathbf A such that Q = Q_{\mathbf A}.

Proof. Since Q is a quadratic form, the map [\cdot, \cdot] : V \times V \to \mathbb K defined by

[\mathbf u,\mathbf v] := \frac 12 (Q(\mathbf u + \mathbf v) - Q(\mathbf u) - Q(\mathbf v))

is symmetric and bilinear. Define \mathbf A = [a_{ij}] by a_{ij} = [\mathbf e_i, \mathbf e_j] so that [\mathbf u, \mathbf v] = \mathbf u^{\mathrm T}\mathbf A \mathbf v. Then Q(\mathbf v) = [\mathbf v, \mathbf v] = \mathbf v^{\mathrm T} \mathbf A \mathbf v = Q_{\mathbf A}(\mathbf v).

In fact, more is true.

Lemma 4. Let V be any vector space over \mathbb K. Then any symmetric and bilinear form [\cdot, \cdot] : V \times V \to \mathbb K induces a unique quadratic form Q : V \to \mathbb K.

Proof. Define Q : V \to \mathbb K by Q(\mathbf v) = [\mathbf v, \mathbf v]. By the symmetry and bilinearity of [\cdot, \cdot],

\begin{aligned} Q(\mathbf u + \mathbf v) &= [\mathbf u + \mathbf v, \mathbf u + \mathbf v] \\ &= [\mathbf u , \mathbf u] + [\mathbf u , \mathbf v] + [\mathbf v , \mathbf u] + [\mathbf v , \mathbf v] \\ &= Q(\mathbf u) + Q(\mathbf v) + 2 [\mathbf u, \mathbf v], \end{aligned}

as required.

This leads us to a rather fascinating discussion on rational trigonometry, which aims to discuss trigonometry and geometry on a much more computationally tractable foundation, introduced by UNSW Professor Norman Wildberger.

For the rest of this post, suppose [\cdot, \cdot] is a symmetric and bilinear form on V. If \mathbb K = \mathbb R, then any inner product is a symmetric and bilinear form on V.

Definition 2. Define the quadrance of \mathbf v \in V by the quadratic form Q(\mathbf v) = [\mathbf v,\mathbf v]. For vectors \mathbf u, \mathbf v \in V with nonzero quadrance, define the cross of the vectors by

\displaystyle c(\mathbf u, \mathbf v) := \frac{[\mathbf u, \mathbf v]^2}{Q(\mathbf u) Q(\mathbf v)}.

Finally, define the spread of the vectors by s := 1 - c.

Remark 1. Three nonzero vectors \mathbf u, \mathbf v, \mathbf w form a triangle precisely when \mathbf u + \mathbf v + \mathbf w = \mathbf 0. Subsequent discussions apply to that of a triangle since Q(-\mathbf w) = Q(\mathbf w).

Theorem 1 (Cross Law). For \mathbf u, \mathbf v \in V,

(Q(\mathbf u) + Q(\mathbf v) - Q(\mathbf u + \mathbf v))^2 = 4 Q(\mathbf u) Q(\mathbf v) c(\mathbf u, \mathbf v).

Proof. The left-hand side simplifies to 4[\mathbf u, \mathbf v]^2.

Theorem 2 (Pythagoras’ Theorem). For \mathbf u, \mathbf v \in V, we have [\mathbf u, \mathbf v] = 0 if and only if

Q(\mathbf u + \mathbf v) = Q(\mathbf u) + Q(\mathbf v).

Theorem 3 (Triple Quad Formula). For \mathbf u, \mathbf v \in V, we have s(\mathbf u, \mathbf v) = 0 if and only if

(Q(\mathbf u) + Q(\mathbf v) - Q(\mathbf u + \mathbf v))^2 = 4 Q(\mathbf u) Q(\mathbf v).

Equivalently,

(Q(\mathbf u) + Q(\mathbf v) + Q(\mathbf u + \mathbf v))^2 = 2(Q(\mathbf u)^2 + Q(\mathbf v)^2 + Q(\mathbf u + \mathbf v)^2).

Proof. The result is immediate from the equivalence s = 1 - c. For the equivalent statement, denote Q_1 = Q(\mathbf u), Q_2 = Q(\mathbf v), Q_3 = Q(\mathbf u + \mathbf v). Then

\begin{aligned} (Q_1 + Q_2 + Q_3)^2 &= (Q_1 + Q_2 - Q_3 + 2 Q_3)^2 \\ &= (Q_1 + Q_2 - Q_3)^2 + 2(Q_1 + Q_2 - Q_3)\cdot 2 Q_3 + 4Q_3^2 \\ &= 4Q_1 Q_2 + 2(Q_1 + Q_2 - Q_3)\cdot 2 Q_3 + 4Q_3^2 \\ &= 2 \cdot 2(Q_1 Q_2 + Q_1Q_3 + Q_2Q_3) \\ &= 2 \cdot ((Q_1 + Q_2 + Q_3)^2 - (Q_1^2 + Q_2^2 + Q_3^2)) \\ &= 2 (Q_1 + Q_2 + Q_3)^2 - 2 (Q_1^2 + Q_2^2 + Q_3^2), \end{aligned}

from which algebra yields the desired result.

Theorem 4 (Spread Law). For \mathbf u, \mathbf v \in V, suppose \mathbf u, \mathbf v, \mathbf u +\mathbf v have nonzero quadrances Q_1, Q_2,Q_3 respectively. Define

\begin{aligned} s_1 &:= s(\mathbf v, \mathbf u + \mathbf v) ,\quad s_2 := s(\mathbf u, \mathbf u + \mathbf v), \quad s_3 := s(\mathbf u, \mathbf v).\end{aligned}

Then \displaystyle \frac{s_1}{Q_1} = \frac{s_2}{Q_2} = \frac{s_3}{Q_3}.

Proof. By the Cross law,

\begin{aligned} (Q_1 + Q_2 - Q_3)^2 &= 4 Q_1 Q_2 (1 - s_3). \end{aligned}

Recalling the proof of Theorem 3,

\begin{aligned} (Q_1 + Q_2 + Q_3)^2 &= (Q_1 + Q_2 - Q_3 + 2 Q_3)^2 \\ &= (Q_1 + Q_2 - Q_3)^2 + 2(Q_1 + Q_2 - Q_3)\cdot 2 Q_3 + 4Q_3^2 \\ &= 4Q_1 Q_2 + 2(Q_1 + Q_2 - Q_3)\cdot 2 Q_3 + 4Q_3^2 - 4 Q_1 Q_2 s_3 \\ &= 2 (Q_1 + Q_2 + Q_3)^2 - 2 (Q_1^2 + Q_2^2 + Q_3^2) - 4 Q_1 Q_2 s_3, \end{aligned}

so that 4 Q_1 Q_2 s_3 = (Q_1 + Q_2 + Q_3)^2-2 (Q_1^2 + Q_2^2 + Q_3^2) . By the symmetry of the right-hand side,

4 Q_1 Q_2 s_3 = 4 Q_2 Q_3 s_1 = 4 Q_3 Q_1 s_2.

Dividing by 4Q_1 Q_2 Q_3 on all sides,

\displaystyle \frac{s_1}{Q_1} = \frac{s_2}{Q_2} = \frac{s_3}{Q_3} =: \frac 1D.

Theorem 5 (Triple Spread Formula). Following the notation in Theorem 4,

\begin{aligned} (s_1 + s_2 + s_3)^2 = 2(s_1^2 + s_2^2 + s_2^2) + 4 s_1 s_2 s_3.\end{aligned}

Proof. For each i, Q_i = D s_i. Substituting into the equivalent of the Cross law,

4 Ds_1 Ds_2 s_3 = (Ds_1 + Ds_2 + Ds_3)^2- 2 (D^2 s_1^2 + D^2 s_2^2 + D^2 s_3^2) .

Dividing by D^2 yields

4 s_1 s_2 s_3 = (s_1 + s_2 + s_3)^2-2 ( s_1^2 + s_2^2 + s_3^2) ,

as required.

For more fascinating details about rational trigonometry, check out Prof Wildberger’s YouTube channel. In particular, we want to use rational trigonometry to create geometry in the next post.

—Joel Kindiak, 20 Mar 25, 1024H

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    […] Last time, we discussed symmetric bilinear forms and its applications into geometry, in particular, to rational trigonometry, which posits a computationally tractable alternative to classical trigonometry. The rational analog of distance would be the quadrance , and the rational analog of angle would be the spread , connected in the case by the equations […]

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