The half-angle formulas are central!

 As the title suggests, the half-angle formulas are central. Even more central than the law of cosines, which is nothing more than the half-angle formulas in disguise. Virtually, every metric relationship characterizing the triangle can be derived from the triangular half-angle formulas. The mind map below can give you an idea of the pivotal role that the half-angle formulas play in relation to the other most important metric relationships in classical geometry. I have also written an essay titled "The Theoretical Importance of Half-Angle Formulas" where you can see the details of the proofs, as well as the new generalizations I have managed to derive from this novel approach.

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Identifying the central theorems of an area is important because it helps streamline the process of understanding it. By knowing the basic principles, you can, with logic and a bit of ingenuity, understand (by proving) the rest of the discipline without having to memorize as much.

Solution to Problem 1315 of Gogeometry

 The following problem was proposed by my esteemed friend, Ángel Mejía. However, this is problem 1315 from Gogeometry. Here I give a trigonometric proof.

The figure shows a triangle $ABC$ with the inscribed circle $O$ ($D$, $E$, and $T$ are the tangency points). $OB$ cuts chord $DE$ and arc $DE$ at $M$ and $F$, respectively. $AF$ and $CF$ cut chord $DE$ at $G$ and $N$, respectively. Prove that $DG = MN$.

Proof. Let's denote $\angle{BAC}=\alpha$, $\angle{ABC}=\beta$, $\angle{ACB}=\gamma$, $\angle{AFM}=\angle{AFO}=\delta$, and $\angle{DFA}=\epsilon$. Applying the sine law in triangles $GFM$ and $DFG$, we obtain

$$GM = FG \cdot \sin{\delta},\tag{1}$$

$$DG = \frac{FG \cdot \sin{\epsilon}}{\sin{\frac12\left(\frac{\pi}{2}-\frac{\beta}{2}\right)}}.\tag{2}$$

Dividing $(1)$ by $(2)$ and then squaring, we get

$$\left(\frac{GM}{DG}\right)^2=\frac{\sin^2{\delta}}{\sin^2{\epsilon}}\cdot{\sin^2{\frac12\left(\frac{\pi}{2}-\frac{\beta}{2}\right)}}=\frac{\sin^2{\delta}}{\sin^2{\epsilon}}\left(\frac{1-\sin{\frac12\beta}}{2}\right).\tag{3}$$

Now, applying the sine law once more, this time in triangles $AFO$ and $ADF$, we obtain

$$\sin{\delta}=\frac{AO\cdot{\sin{\frac12(\alpha+\beta)}}}{AF},\tag{4}$$

$$\sin{\epsilon}=\frac{AD\cdot{\sin{\frac14(3\pi+\beta})}}{AF}.\tag{5}$$

Dividing $(4)$ by $(5)$ and then squaring, we have

$$\frac{\sin^2{\delta}}{\sin^2{\epsilon}}=\frac{\sec^2{\frac12\alpha}\cos^2{\frac12\gamma}}{\sin^2{\frac14(3\pi+\beta})}=\sec^2{\frac12\alpha}\cos^2{\frac12\gamma}\left(\frac{2}{1-\sin{\frac12\beta}}\right).\tag{6}$$

Substituting $(6)$ in $(3)$ and taking square roots,

$$\frac{GM}{DG}=\frac{\cos{\frac12\gamma}}{\cos{\frac12\alpha}}.$$

Analogously,

$$\frac{MN}{EN}=\frac{\cos{\frac12\alpha}}{\cos{\frac12\gamma}}.$$

This implies $\frac{GM}{DG}=\frac{EN}{MN}$ and, since $DM=ME$, we conclude that $DG=MN$.

$\square$

Another Identity on Triangle Areas Arising from Reflection

In a previous post, we proved an identity about triangle areas arising from reflection. In this occasion, we prove another interesting identity associated with triangle areas.

Theorem. Let $ABC$ be a triangle. Denote $D$, $E$ and $F$ arbitrary points on sides $BC$, $AC$ and $AB$, respectively. Let $A'$ be the reflection of $A$ with respect to $D$. Define $B'$ and $C'$ similarly. Denote $A''$ the reflection of $A'$ with respect to the midpoint of $BC$. Define $B''$ and $C''$ similarly. Then 

$$[A'B'C']-[A''B''C''] = 3[ABC].$$

The brackets $[\, ]$ represent the area of the enclosed figure.

Lemma 1. $[A'B'C]=[ABA''B'']$, $[A'C'B]=[ACA''C'']$ and $[B'C'A]=[BCB''C'']$.

Proof. Observe that the diagonals of $BCB'C'$ and $BCEF$ are in a 2:1 ratio, and the angle between them remains unchanged. Hence, $[BCB'C']=4[BCEF]$. This is easily deduced from the formula for the area of a quadrilateral $K=\frac12pq\sin{\theta}$, where the lengths of the diagonals are $p$ and $q$, and the angle between them is $\theta$.

Also, observe that $AF$ and $BF$ are medians of triangles $\triangle{ACC'}$ and $\triangle{BCC'}$, respectively. It follows that $[ACF]=[AC'F]$ and $[BCF]=[BC'F]$, and then $[ABC]=[ABC']$. Analogously, $[ABC]=[ACB']=[BCA']$. Now, it turns out that

$$[AB'C']=[BCB'C']-3[ABC]=4[BCEF]-3[ABC]=4[ABC]-4[AEF]-3[ABC]=[ABC]-4[AEF].\tag{1}$$

Let $H$ and $I$ be the midpoints of $AC$ and $AB$, respectively. Then

$$[ABC]=4[AHI]=4[AEF]+4[HEFI].\tag{2}$$

From $(1)$ and $(2)$ follows that $[AB'C']=4[HEFI]$. Now, observe that $BC$, $BB''$, and $CC''$ are the homothetic images with a scale factor of 2 of $IH$, $EH$, and $FI$ with respect to $A$, $B'$, and $C'$, respectively. It follows that $BCC''B''$ and $HEFI$ are homothetic with $[BCC''B'']=4[HEFI]$, therefore,

$$[AB'C']=[BCC''B''].$$

$\square$

A similar reasoning must show that $[A'B'C]=[ABB''A'']$, and for the case of $\triangle{A'C'B}$ and $ACA''C''$ where $F$ and $D$ are on opposite sides of the line $GI$, where $G$ is the midpoint of the side $BC$, only a minor adjustment will be required.

$\color{blue}{[A'B'C]=[ABA''B'']}$, $\color{red}{[A'C'B]=[ACA''C'']}$ and $\color{green}{[B'C'A]=[BCB''C'']}$.

Remark. Notice lemma 1 gives us an alternative way to construct a triangle with an area equal to a given quadrilateral, as long as it is not a rectangle or a square.

Back to the main problem

Notice that 

$$[A'B'C']=[A'C'B'C]-[A'B'C]=4[ABC]+[AB'C']+[BA'C']-[A'B'C].$$

But $[AB'C']=[ABC]-4[AEF]$ and similarly $[BA'C']=[ABC]-4[BDF]$ and $[A'B'C]=4[CDE]-[ABC]$, so

$$[A'B'C']= 7[ABC]-4[AEF]-4[BDF]-4[CDE].\tag{3}$$

As $HIFE$ and $BCC''B''$ are inversely homothetic with scale factor $-2$, then so are segments $EF$ and $B''C''$ and similarly for $ED$ and $A''C''$ and $FD$ and $A''C''$, meaning that triangles $\triangle{DEF}$ and $\triangle{A''B''C''}$ are also inversely homothetic with factor $-2$ and then

$$[A''B''C'']=4[DEF]=4[ABC]-4[AEF]-4[BDF]-4[CDE].\tag{4}$$

Substracting $(4)$ from $(3)$ we get 

$$[A'B'C']-[A''B''C'']=3[ABC].$$

$\square$

A curious family of integrals that give rational multiples of $\pi$

 I have noticed experimentally that:

$$\int_0^1 \frac{\color{red}{x}}{x^4+2x^3+2x^2-2x+1} \,dx=\color{blue}{\frac{\pi}{8}},\tag{1}$$

$$\int_0^1 \frac{\color{red}{1-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx=\color{blue}{\frac{\pi}{4}},\tag{2}$$

$$\int_0^1 \frac{\color{red}{1+x-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx =\color{blue}{\frac{3\pi}{8}}.\tag{3}$$

So slight variations in the numerator always seem to produce something like $n\pi$, where $n$ is a rational number.

At MathSE I have asked what the exact relationship is between $n$ and the numerator of the integrand, to which Quanto has responded with a general formula:

$$\int_{0}^{1} \frac{ax^2 +b x + c}{x^4+2x^3+2x^2-2x+1} \, dx=\frac\pi8\color{green}{(c+b-a)}+\frac\pi{3\sqrt3}\color{green}{(a+c)}.\tag{4}$$

However, is it necessary for the denominator to remain fixed? Not really. The following integral is formula $(34)$ in this list of $\pi$ formulas:

$$\int_0^1 \frac{\color{red}{16x-16}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\pi}.\tag{5}$$

Notice that the denominator is different. But again, a slight variation in the numerator and it still produces something like $n\pi$:

$$\int_0^1 \frac{\color{red}{x^2-x-1}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\frac{3\pi}{16}}.\tag{6}$$

More generally, 

$$\int_{0}^{1} \frac{ax^2+bx+c}{x^4-2x^3+4x-4}\,dx=\frac{\pi}{16}\color{green}{(2a-c)}+\frac{\ln{(3-\sqrt8)}}{\sqrt32}\color{green}{(b+c)}+\frac{\ln{(3-\sqrt8)}}{\sqrt8}\color{green}{a}.\tag{7}$$

From $(7)$, we can deduce that the integral will yield rational multiples of $\pi$ when $b$ and $c$ are opposite to each other and $a=0$. However, this formula is still far from being the ultimate generalization since it does not take into account the coefficients of the denominator. The fact that the denominator is not the same in $(4)$ and $(7)$ suggests that a further generalization is possible.

A more intimidating integral

Doing some arithmetic with the integrands, we can obtain more intimidating integrands that still yield $\pi$. The following one comes from adding integrands (2) and (5) with different denominators:

$$\int_{0}^{1} \frac{2(1-x)(x^5-5x^4-10x^3-4x^2+8x-8)}{x^8-2x^6-2x^5+9x^4-2x^3-16x^2+12x-4}=\pi.$$

Addendum. I wonder if it is possible to characterize the integrand $\frac{P(x)}{Q(x)}$ in such a way that by simple inspection we can say $I=n\pi$? Or in other words, what should be the relationship between the coefficients of the numerator and the denominator for the integral to yield $n\pi$?

Sine half-angle substitution

 If you are a student of integral calculus, it is highly likely that you have come across or will come across the famous Weierstrass substitution, which is very useful for converting rational expressions involving trigonometric functions into ordinary rational expressions involving $t$, where $t=\tan{\frac12{x}}$. The general transformation formula is as follows:

$$\int f(\sin{x},\cos{x})\,dx =  \int f\left(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2}\right)\frac{2dt}{1+t^2}.$$

In this note, we introduce another substitution as a companion to the Weierstrass substitution that can transform certain rational expressions of trigonometric functions into simpler ordinary rational expressions than the Weierstrass substitution. For instance, it would be useful when a common linear factor of $\sin{x}$ appears in the numerator or denominator of the integrand. The general transformation formula is given by:

$$\int f(\sin{x},\cos{x})\,dx =  \int f(2\sqrt{s-s^2}, 1-2s)\frac{ds}{\sqrt{s-s^2}}.$$

Here $s=\sin^2{\frac12x}$. 

Derivation

Using the double-angle formulas and the Pythagorean identity, one gets
$$\sin{x}=2\sin{\frac12x}\cos{\frac12x}=2\sin{\frac12x}\sqrt{1-\sin^2{\frac12x}}=2\sqrt{\sin^2{\frac12x}-\sin^4{\frac12x}}=2\sqrt{s-s^2},$$
$$\cos{x}=1-2\sin^2{\frac12x}=1-2s.$$
Finally, since $s=\sin^2{\frac12x}$, differentiation rules imply
$$ds=\sin{\frac12x}\cos{\frac12x}\,dx=\frac{\sin{x}}{2}\,dx,$$
and thus,
$$dx=\frac{ds}{\sqrt{s-s^2}}.$$

Example 1

By applying the sine half-angle substitution and simplifying,

$$\int \frac{\sin{x}}{\sin^2{x}+2\cos{x}}\,dx=\int \frac{1}{1-2s^2}\,ds.$$

Using standard integrals,

$$\int \frac{1}{1-2s^2}\,ds=\frac{\tanh^{-1}{(\sqrt{2}s)}}{\sqrt{2}}+C=\frac{\tanh^{-1}{(\sqrt{2}\sin^2{\frac12x})}}{\sqrt{2}}+C.$$

The advantage of this substitution is evident when comparing it to the solution provided in this integral calculator (which solution, by the way, is equivalent to the one given here) or when using the Weierstrass substitution.

Example 2 

By applying the sine half-angle substitution,

$$\int \frac{1}{\sin^3{x}}\,dx=\frac18\int \frac{1}{(s-s^2)^2}\,ds.$$

Using partial fraction decomposition, 

$$\frac18\int \frac{1}{(s-s^2)^2}\,ds=\frac18\int \left(\frac{1}{s}+\frac{1}{s^2}-\frac{2}{s-1}+\frac{1}{(s-1)^2}\right)\,ds=\frac14\ln{\left|\frac{s}{1-s}\right|}+\frac18\left(\frac{1}{1-s}-\frac{1}{s}\right)+C.$$

Substituting $s$ by $\sin^2{\frac12x}$, we have

$$\frac18\int \frac{1}{(s-s^2)^2}\,ds=\frac14\ln{\left|\tan^2{\frac12x}\right|}+\frac18\left(\frac{1}{\cos^2{\frac12x}}-\frac{1}{\sin^2{\frac12x}}\right)+C=\frac12\left(\ln{|\tan{\frac12x}|}-\frac{\cos{x}}{\sin^2{x}}\right)+C.$$

A solution using integration by parts is given at Integrals For You. 

A Generalization of the Law of Sines

 The following is an extension of the law of sines for cyclic quadrilaterals that comes to accompany the generalization of Mollweide's formula (rather Newton's) and the generalization of the law of tangent.

Generalization. Consider a cyclic quadrilateral, $ABCD$, with side lengths $AB=a$, $BC=b$, $CD=c$, and $DA=d$. Let $\angle{DAB}=\alpha$, $\angle{ABD}=\beta$, $\angle{BCD}=\gamma$, and $\angle{CDA}=\delta$. Then the following identity holds

$$\frac{ab+cd}{\sin{\alpha}}=\frac{ad+bc}{\sin{\beta}}=\frac{ab+cd}{\sin{\gamma}}=\frac{ad+bc}{\sin{\delta}}.$$

At first, I was reluctant to publish this result because the proof is very straightforward (hence, I leave it as an exercise to the reader). However, I have shared it on my social media platforms (see here, here and here) and the audience's response has been more favorable than my generalizations of Mollweide's formula and the law of tangents combined.

Related material

Generalizing Lami's theorem

A new proof of angle sum identity for the sine

Several interesting proofs of angle sum identity are given on MathSE. In this note, I will provide another proof that is possible because, although traditionally presented in textbooks as a consequence of the angle sum identity $\sin{(x+y)}$, the double-angle formula for the sine, $\sin{(2x)}=2\sin{(x)}\cos{(x)}$, can be derived independently of it (see the wonderful proof without words of "Start wearing purple"). Then, as Blue has pointed out, the fact that supplementary angles have the same sine is an easy consequence of the double-angle formula for the sine.

The following proof is valid for $0\le\alpha\le\pi$, $0\le\beta\le\dfrac{\pi}{2}$, $0\le\alpha+\beta\le\pi$.

Theorem.  The sum identity for sine states that

$$\sin{(\alpha+\beta)}=\sin{(\alpha)}\cos{(\beta)}+\cos{(\alpha)}\sin{(\beta)}.\tag{1}$$

Proof. Suppose $\triangle{ABC}$ is a triangle with sides $\overline{BC}=a$, $\overline{AC}=b$ and $\overline{AB}=c$. Let $\angle{BAC}=\alpha$, $\angle{CBA}=\beta$ and $\angle{ACB}=\gamma$. Recalling that the area of a triangle can be expressed as $\Delta=\frac12bc\sin{(\alpha)}$ and substituting from the cosine rule on the right-hand side of $(1)$, we have  

\[\begin{aligned} \sin{(\alpha)}\cos{(\beta)}+\cos{(\alpha)}\sin{(\beta)} &= \left(\frac{2\Delta}{bc}\right)\cos{(\beta)}+\left(\frac{2\Delta}{ac}\right)\cos{(\alpha)}\\&=\frac{2\Delta}{c}\left(\frac{\cos{(\beta)}}{b}+\frac{\cos{(\alpha)}}{a}\right)\\ &=\frac{2\Delta}{c}\left(\frac{a^2+c^2-b^2}{2abc}+\frac{b^2+c^2-a^2}{2abc}\right)\\&=\frac{2\Delta}{ab}\\&=\sin{(\gamma)}\\&=\sin{(\pi-(\alpha+\beta))}\\&=\sin{2\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)}\\&=2\color{red}{\sin{\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)}}\color{blue}{\cos{\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)}}\\&=2\color{red}{\cos{\left(\frac{\alpha+\beta}{2}\right)}}\color{blue}{\sin{\left(\frac{\alpha+\beta}{2}\right)}}\\&= \sin{\left(\alpha+\beta\right)}.\end{aligned}\]

$\square$

Breve discurso ante una multitud de estudiantes: el enfoque de medio ángulo (EMA)

Hace dos semanas, recibí una invitación para dar un breve discurso a estudiantes de secundaria acerca de mis más recientes descubrimientos en geometría plana. A continuación, les comparto lo que dije:

«Hola, yo soy Emmanuel y he venido literalmente a presumir, ¿qué les parece? Pero antes de empezar quiero hacerles la siguiente pregunta: ¿saben ustedes qué es el informe PISA? Pues es un estudio llevado a cabo a nivel mundial que mide el rendimiento académico de los alumnos de 15 años en matemáticas, ciencia y lectura comprensiva. Si usted quiere encontrar a República Dominicana en la larga lista de países que participan en este estudio, usted solo debe dirigirse al final de la lista, ¡qué mal! Aunque cuando cuento esto a mis estudiantes muchos se echan a reír, es una situación para llorar. ¡Esto es peor que la barrida que nos dieron en el Clásico Mundial de Baseball!

Pero…a pesar de estos resultados tan alarmantes, aunque usted no lo crea, en República Dominicana se ha descubierto un nuevo enfoque (al que yo llamo EMA) que me ha permitido generalizar una fórmula de más de 900 años: la ley de la tangente. Lo que quería decir es que luego de casi un año de espera, desempolvando libros y preguntando a matemáticos de todas partes del mundo, recientemente se me informó que mi generalización será publicada en la prestigiosa revista americana Mathematics Magazine. Pero la cosa no se queda ahí, este nuevo enfoque me ha permitido también generalizar las fórmulas de medio ángulo, la identidad pitagórica, la ley de senos (¡me honra que el gran John Baez haya compartido mi generalización!), el teorema de Lamy y la fórmula de Newton, además de poder derivar una plétora de fórmulas y teoremas bien conocidos en geometría plana.

Pero, ¿qué es una generalización? Una generalización es, en palabras llanas, ampliar el alcance de un teorema. Grandes matemáticos tienen sus propias generalizaciones. ¿Qué creen ustedes que es el binomio de Newton o la famosa fórmula de Euler? ¡Generalizaciones! Por cierto, no me pregunten las horas que pasé desarrollando todo esto.

Para finalizar quiero citar al Dr. James Cook, de la Universidad de Alabama, quien escribió lo siguiente sobre mi enfoque:

Creo que has presentado un caso convincente de que estas fórmulas son bastante básicas. Por supuesto, sospecho que podría derivarse casi todo partiendo de la ley de los cosenos. ¿Qué es la ley de cosenos sino el corazón del producto escalar? Y, ¿qué es el producto escalar? Es la encapsulación algebraica del ángulo. Como mínimo, esto debería aparecer como problemas o una sección de tema adicional en los textos de trigonometría. Parece que esto sería excelente para un curso de honor de escuela de verano para dotados en matemáticas. El hecho de que no se enseñe podría aprovecharse para permitir que los estudiantes lo descubran. »

The Half-Angle Formulas: A Powerful Tool for Trigonometry Students

 Let me tell you about an exciting discovery I made while exploring the theoretical importance of half-angle formulas in trigonometry. These formulas, often overlooked and underappreciated in comparison to the more well-known laws of sines, cosines, and tangents, have a wealth of untapped potential waiting to be discovered.

The standard half-angle formulas

I'm not talking about the standard version of the half-angle formulas that appear in most introductory trigonometry textbooks. No, I'm referring to a version that relates the sides of a triangle, the perimeter, and its angles. These formulas have been largely ignored by the mathematical community, with even the almighty Wikipedia lacking an article on them.

The half-angle formulas for a triangle

My journey began when I stumbled upon (1) while attempting to prove the law of cosines by contradiction. I later discovered that this formula was already known, but that didn't stop me from exploring its potential. I went on to generalize the formula and discovered (2), which applies to cyclic quadrilaterals. When I found out that (2) was also already known, I refused to give up and continued to push the limits of these formulas.

The half-angle formulas for a cyclic quadrilateral

What I discovered next was truly mind-blowing. I managed to generalize (2) even further to (3), which applies to general quadrilaterals. Surprisingly, (3) also generalizes the Pythagorean trigonometric identity! These formulas not only provide a new framework for the Heron-Brahmagupta-Bretschneider development (see Two Identities and their Consequences, pp. 5), but they also have the potential to derive a plethora of other formulas and theorems, including the law of cosines, the law of sines, the law of tangents, and Stewart's theorem.

The half-angle formulas for a general quadrilateral

And it doesn't stop there. Using these half-angle formulas, I also managed to derive a generalization of Mollweide's formula and a generalization of the law of tangents. The possibilities are endless, and I'm only scratching the surface of what these formulas can do.

But what excites me the most about these formulas is their potential to revolutionize the way we teach trigonometry to high school students. By introducing the half-angle formulas, we can help students better understand the development of the formulas that derive from them. This is where the half-angle formulas truly shine and can make a significant impact on math education.

In conclusion, the theoretical importance of the half-angle formulas cannot be overstated. They have the potential to unlock a whole new world of mathematical discoveries and have already proven to be a powerful tool for deriving a wide range of formulas and theorems. And with the right approach, they can also be a game-changer for teaching trigonometry to the next generation of students. The possibilities are truly endless, and I can't wait to see what else these formulas have in store for us.

See also The theoretical importance of the half-angle formulas.