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Unlike the laws of sines, cosines and tangents, which are very well known, the half-angle formulas seem (although they appear timidly in the mathematical literature) not to enjoy the same popularity. Thus, while there are entire chapters devoted to the law of sines, cosines, and tangents and their applications, there is not even a Wikipedia article on half-angle formulas. Right now you may be imagining this version of the half angle formulas

$$\sin{\frac{\alpha}{2}}=\pm\sqrt{\frac{1-\cos{\alpha}}{2}}\qquad\qquad\cos{\frac{\alpha}{2}}=\pm\sqrt{\frac{1+\cos{\alpha}}{2}},$$

that do appear in the textbooks of the first trigonometry courses (at least with the one I studied). But actually I mean these

$$\sin^2{\frac{\alpha}{2}} = \frac{(s-b)(s-c)}{bc}\qquad\qquad\cos^2{\frac{\alpha}{2}}= \frac{s(s-a)}{bc},\tag{1}\label{1}$$

where $a$, $b$, and $c$ are the sides of a triangle, $\alpha$ is the angle opposite side $a$, and $s$ is the semiperimeter. I found the furthest reference to these formulas in a conversation posted online between Conway and Doyle, where Conway uses them to prove Heron's formula and later claims to have taken it from a sequel by Casey.

I discovered \eqref{1} independently trying to prove the law of cosines by contradiction. When I realized that they were known, I tried to generalize them and I got this

$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad+bc}\qquad\qquad \cos^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)}{ad+bc},\tag{2}\label{2}$$

where $a$, $b$, $c$ and $d$ are the sides of a cyclic quadrilateral, $s$ is semiperimeter and $\angle{DAB}=\alpha$.

Before discovering the conversation between Conway and Doyle, I had been excited that I had found an original proof of Heron's formula using \eqref{1}. When I found \eqref{2}, I thought that by analogous reasoning I could prove Brahmagupta's formula. So it was. But in a geometry forum someone referred me to an ancient Greek book that contained \eqref{2}. I then sent my proof of Brahmagupta's formula to Martin Josefsson who referred me to Casey's book "A Treatise On Plane Trigonometry" where my proof already appeared. But I didn't give up and tried to generalize \eqref{2} getting this

$$ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}=(s-a)(s-d)\qquad\qquad bc\sin^2{\frac{\gamma}{2}}+ad\cos^2{\frac{\alpha}{2}}=(s-b)(s-c),\tag{3}\label{3}$$

where $a$, $b$, $c$ and $d$ are the sides of a general quadrilateral, $s$ is semiperimeter, $\angle{DAB}=\alpha$ and $\angle{BCD}=\gamma$. Surprisingly, $(3)$ also generalizes the Pythagorean identity. Take a look at GeoDom - A generalization of the Pythagorean trigonometric identity.

Bretschneider's formula is known to be a generalization of Heron's and Brahmagupta's formulas. Naturally, I wondered if I could generalize Casey's proof of Brahmagupta's formula using \eqref{3} and thus derive Bretschneider's formula. And I did it. I sent my formulas in \eqref{3} and my proof of the Bretschneider's formula to Josefsson (among many other mathematicians) and he told me this:

"I like your paper, especially how you put these important formulas in a single framwork. I cannot say that I remember seeing the identities (4) and (5) anywhere else before."

Where identities (4) and (5) are the identities in \eqref{3} in this post.

And then he said:

"Even though much has already been written about these formulas, the ideas for proving Bretschneider' formula and the area of a bicentric quadrilateral are novel as far as I know. I hope you get your paper published."

I decided to write an article about these formulas called "Two Identities and their Consequences" which was published in MATINF .

In almost three years of exploring possible applications of \eqref{1}, \eqref{2}, \eqref{3}, this is what I have found:

Using \eqref{1}, \eqref{2} we can also derive (you can see most of the proofs at GeoDom - Proofs and applications of two well-known formulae involving sine, cosine and the semiperimeter of a triangle):

  • The law of cosines
  • The law of sines
  • The law of tangents
  • Stewart's theorem
  • Compound angle formulas
  • Mollweide's formula
  • The product $AI\cdot{BI}\cdot{CI}$
  • The bisector length formula
  • Mahavira's formulas
  • Zelich's lemma
  • Euler's triangle inequality
  • Yun’s Inequality (Josefsson)

Other unnamed identities and inequalities:

  • $\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=1$
  • $r=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}$
  • $s=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$
  • $\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}\le\frac{1}{8}$
  • Triangle with $\tan{\frac{\alpha}{2}}=\frac{a}{b+c}$ (see Graubner's solution)
  • I could go on and on and on…

The formulas \eqref{1}, \eqref{2}, \eqref{3} explain the Heron–Brahmagupta–Bretschneider development better than I have seen anywhere else. This made me wonder what would happen if I analogously applied the half-angle formulas to formulas where half-angles explicitly appeared, such as Mollweide's (rather Newton's) formula or the law of tangents. This is how these two generalizations arose:

When questioning Martin Josefsson about the originality of these generalizations, this is what he said:

"As far as I can recall, I have not seen any of them, at least not in modern books or papers, and even if some of them where to be found in an old text, they are at least not well known, and deserve to be wider known."

Apart from the proof of the Bretschneider's formula, I haven't found any other applications for \eqref{3}.

Interestingly, half angles seem to be everywhere: from circle angle theorems to the Weierstrass Substitution technique in Integral Calculus. Even when Viète derived his formula for $\pi$ using an infinite product, he started by writing $\sin{x}=2\sin{\frac12x}\cos{\frac12x}$.

Edited question: Doesn't the fact that we can derive all the aforementioned classical theorems from \eqref{1} make \eqref{1} fundamental to Euclidean geometry? Are there other aspects to take into consideration? If so, what would these aspects be?

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    $\begingroup$ This is more of a speech than a question. :) The question part is opinion-based, which may attract close-votes, but nevertheless ... While I appreciate the attention you've brought to $(1,2,3)$, I believe (and am disappointed) that geometric trig has been mostly driven-out of secondary school curricula. Nowadays, outside of a few professional journals, it's at best the stuff of enthusiast forums (like Art of Problem Solving) and olympiad prep. If you want to see this in a textbook, you may have to write your own. :) (Some books of this flavor may exist, so you could contact their authors.) $\endgroup$
    – Blue
    May 22, 2022 at 21:51
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    $\begingroup$ Simulposted to MO, mathoverflow.net/questions/423102/… $\endgroup$ May 23, 2022 at 4:15
  • $\begingroup$ I think this is somehow related to my question: math.stackexchange.com/questions/572585/… $\endgroup$ Jun 5, 2022 at 14:40
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    $\begingroup$ The goal of Math SE is to create a repository of narrowly focused questions which admit authoritative, objective answers. It is not a discussion forum, not a platform for the dissemination of new work (there are better places for those activities, e.g. reddit, a personal blog, etc). This question seems quite broad, appears to be looking to start a discussion, and doesn't seem to admit an authoritative, objective answer. As such, I do not think that it is a very good fit for this site. $\endgroup$
    – Xander Henderson
    Jun 29, 2022 at 13:15
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    $\begingroup$ Moreover, you have repeatedly edited the question. Editing a question bumps it to the top of the front page, which takes visibility away from other questions, and it creates a kind of "moving target" for people who might be interested in answering the question. Please try to avoid making a large number of incremental edits. $\endgroup$
    – Xander Henderson
    Jun 29, 2022 at 13:16

1 Answer 1

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I would argue (admittedly somewhat on a tangent) such half-angle formulas provide a elementary means to prove the fundamental theorem of algebra (a proof accessible to a high-school audience). That is, closely related to the half-angle formulas is an explicit formula for the square root of a complex number $$(x+yi)^{\frac{1}{2}}=\left(\frac{x+(x^2+y^2)^{\frac{1}{2}}}{2}\right)^{\frac{1}{2}}+\text{sgn}(y)\left(\frac{-x+(x^2+y^2)^{\frac{1}{2}}}{2}\right)^{\frac{1}{2}}i.\qquad (*)$$ The existence of such a square root can be used to show that the polynomial $z^n-i$ has roots for arbitrary $n\in \mathbb{N}_0$. From there, it is easy to proceed to a proof to the full fundamental theory of algebra.

To clarify the steps of the proof: let's first show that the equations $z^{2^k}=i$ has a solution, for arbitrary $k\in \mathbb{N}$ and by method of induction on $k$. The proof is trivial for $k=0$. If the statement is proven up to and including given $k\in \mathbb{N}$ and if $z_{k}=x+iy$ is a solution of $z^{2^k}= i$, then $z_{k+1}:=z_{k}^{1/2}$ -given by formula $(*)$ above- solves $(z_{k+1})^{2^{k+1}}= i$. To show that the equation $z^n=i$ has a solution (for arbitrary $n\in \mathbb{N}_0$), decompose $n=m\cdot 2^k$ where $m$ is odd and $k$ integer. If $m=1\text{ mod }4$, then (sticking to the aforementioned definition of $z_k$) $(z_{k})^n=\left(z_{k}^{2^k}\right)^m=i^m=i$, else $(z_{k}^3)^n=\left(z_{k}^{2^k}\right)^{3m}=i^{3m}=i$.

To proceed to the proof of the fundamental Theorem of algebra, suppose $p: \mathbb{C} \to \mathbb{C}$ were a complex-coefficient polynomial of degree $d>0$ without a root. Define $q=|p|^2=\overline{p} \cdot p$. Since $q$ is continuous and $\exists R>0$ s.t. $q|_{\mathbb{C}\setminus D(0,R)}>q(0)$, $q$ must reach a global minimum at some point $z_0\in \mathbb{C}$. WLOG assume $z_0=0$. There is a certain $n\in \{1,...,d\}$ so that (for small $z$) $$q(z)=|p(0)|^2+2 Re\left(\underbrace{\overline{p(0)}\frac{p^{(n)}(0)}{n!}}_{=:\alpha} z^n\right)+O(|z|^{n+1}),\qquad \alpha\neq 0.$$ At least one of the four complex numbers $\alpha, i\alpha,-\alpha,-i\alpha$ has a negative real part. Since the equation $z^n=i$ has a solution, the equations $z^n=-1$ and $z^{n}=-i$ have solutions too, hence there exists a $\xi\in \mathbb{C}$ s.t. $Re(\alpha \xi^n)<0$. Then, for small $t\in (0,+\infty)$, $$q(t\xi)=|p(0)|^2+2Re(\alpha \xi^n) t^n+O(t^{n+1})$$ which is smaller than $q(0)$ for sufficiently small $t$, in contradiction with the assumption that $q$ reached a global minimum at $z=0$.

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