sábado, 20 de junio de 2020

Proofs and applications of two well-known formulae involving sine, cosine and the semiperimeter of a triangle

In Cut-the-knot's Relations between various elements of a triangle, the formulae 

$$\sin^2{\frac{\gamma}{2}} = \frac{(s-a)(s-b)}{ab}\quad and\quad\cos^2{\frac{\gamma}{2}}= \frac{s(s-c)}{ab}$$ 

are derived using Heron's formula. Here we give an alternative proof without using Heron's Formula and we demonstrate several well-known theorems based on these formulae as a sample of its power. We will be using standard notation: $BC=a$, $AC=b$, $AB=c$, $\Delta$ for the area, $s$ for the semiperimeter, $R$ for the circumradius and $r$ for the inradius. Let $D$, $E$ and $F$ be the contact points of the incircle with $AC$, $AB$ and $BC$, respectively. Also, let $AE=AD=x$; $BE=BF=y$; $CD=CF=z$.



Notice that $\frac{\cot{\frac{\gamma}{2}}}{s-c} = \frac{1}{r}$. Also, We know $Δ = rs$ and $Δ =\frac {ab\sin{\gamma}}{2}$, hence

$$\frac{\cot{\frac{\gamma}{2}}}{s-c}=\frac{1}{r}=\frac{s}{\Delta}=\frac{2s}{ab\sin{\gamma}}$$

But, $\sin{\gamma} = 2\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}$, so 
 
$$\frac{\cos{\frac{\gamma}{2}}}{\sin{\frac{\gamma}{2}}}\cdot{\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}} = \frac{s(s-c)}{ab}$$

from which we get $\cos^2{\frac{\gamma}{2}} = \frac{s(s-c)}{ab}$.
 
The other formula can be obtained replacing $\cos^2{\frac{\gamma}{2}}$ by $1 - \sin^2{\frac{\gamma}{2}}$. Indeed, 

$$\begin{aligned} 1-\sin^2{\frac{\gamma}{2}} &= \frac{s(s-c)}{ab} \\ \sin^2{\frac{\gamma}{2}} &= 1-\frac{s(s-c)}{ab} \\  &= \frac{\left(ab-s(s-c)\right)}{ab} \\ &= \frac{\left((y+z)(x+z)-(x+y+z)(z)\right)}{ab}\\ &= \frac{xy}{ab} \\ &=\frac{(s-a)(s-b)}{ab} \end{aligned}$$

$\square$

A proof of Heron's Formula
Making use of the formulae proven above and the double angle identity for sine we have

$$\sin{\gamma}=2\sqrt{\frac{s(s-c)}{ab}}\sqrt{\frac{(s-a)(s-b)}{ab}}=2\frac{\sqrt{s(s-a)(s-b)(s-c)}}{ab}$$

Since $\Delta=\frac{ab\sin{\gamma}}{2}$, it follows 

$$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$

$\square$

A proof of the Law of Cosines
Since $(s-a)=x$, $(s-b)=y$ and $(s-c)=z$, then the following identity holds:
$$ab\cos{\gamma}=ab\cos^2{\frac{\gamma}{2}}-ab\sin^2{\frac{\gamma}{2}}=s(s-c)-(s-a)(s-b)$$
Substituting and multiplying by 4, 
$$4ab\cos{\gamma}=(a+b+c)(a+b-c)-(b+c-a)(a+c-b)$$
Simplifying,
$$2ab\cos{\gamma}=a^2+b^2-c^2$$
$\square$

A similar reasoning must show that $a^2=b^2+c^2-2bc\cos{\alpha}$ and $b^2=a^2+c^2-2ac\cos{\beta}$.


Proofs for some trigonometric identities associated to a triangle
a) $\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=1$.

As a consequence of the formulae proven at the beginning of the note,

$$\tan{\frac{\alpha}{2}}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}, \quad\tan{\frac{\beta}{2}}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\quad and \quad \tan{\frac{\gamma}{2}}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$$

 So, by canceling and simplifying you get

$$\begin{aligned}\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}} &=\frac{s-c}{s}+\frac{s-b}{s}+\frac{s-a}{s}\\ &=\frac{z+y+x}{s}\\ &=\frac{s}{s}=1\end{aligned}$$

$\square$

b) $r=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}$.

We make use of the well-known relationship $abc=4R\Delta$ (see here for a proof) and Heron's Formula.

$$\begin{aligned}r&=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}\\ &=4R\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{(s-a)(s-c)}{ac}}\sqrt{\frac{(s-a)(s-b)}{ab}}\\&=4R\sqrt{\frac{(s-a)^2(s-b)^2(s-c)^2}{a^2b^2c^2}}\\&=4R\sqrt{\frac{\frac{\Delta^4}{s^2}}{a^2b^2c^2}}\\&=4R\sqrt{\frac{\frac{\Delta^4}{s^2}}{16R^2\Delta^2}}\\&=\frac{\Delta}{s}\\&=\frac{rs}{s}\\&=r\end{aligned}$$

$\square$

c) $s=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$.

$$\begin{aligned}s&=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}\\ &=4R\sqrt{\frac{s(s-a)}{bc}}\sqrt{\frac{s(s-b)}{ac}}\sqrt{\frac{s(s-c)}{ab}}\\&=4R\sqrt{\frac{s^2\Delta^2}{a^2b^2c^2}}\\&=4R\frac{s\Delta}{abc}\\&=4R\frac{s\Delta}{4R\Delta}\\&=s\end{aligned}$$

$\square$


Consequently, the following relationship also holds

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=\frac{r}{s}$$

or

$$\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\frac{s}{r}$$

d) $\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\cot{\frac{\alpha}{2}}+\cot{\frac{\beta}{2}}+\cot{\frac{\gamma}{2}}$.

To prove the above identity we will show that the right hand side equals $\frac{s}{r}$.

$$\begin{aligned}\frac{s}{r}&= \cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\cot{\frac{\alpha}{2}}+\cot{\frac{\beta}{2}}+\cot{\frac{\gamma}{2}}\\&=\frac{\sqrt{s(s-a)}}{\sqrt{(s-b)(s-c)}}+\frac{\sqrt{s(s-b)}}{\sqrt{(s-a)(s-c)}}+\frac{\sqrt{s(s-c)}}{\sqrt{(s-a)(s-b)}}\\&=\frac{\Delta(s-a)+\Delta(s-b)+\Delta(s-c)}{(s-a)(s-b)(s-c)}\\&=\frac{\Delta(x+y+z)}{xyz}\\&=\Delta\frac{s^2}{\Delta^2}\\&=\frac{s^2}{rs}\\&=\frac{s}{r}\end{aligned}$$

$\square$


We invite the reader to prove the following identity (possibly new) on their own.
$$\frac{\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}+\cot{\frac{\alpha}{2}}\cot{\frac{\gamma}{2}}+\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}}{(s-a)(s-b)+(s-a)(s-c)+(s-b)(s-c)}=\frac{1}{r^2}$$

Two similar formulae (possibly new) in a cyclic quadrilateral
Similar formulae can be derived from a cyclic quadrilateral. Let $ABCD$ be a cyclic quadrilateral with $AB=a$, $BC=b$, $CD=c$, $DA=d$ and $s=\frac{a+b+c+d}{2}$. If $\angle{BAD}=\alpha$, then


$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad+bc}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)}{ad+bc}$$

First we will find an expression for $\cos{\alpha}$ in terms of $a$, $b$, $c$ and $d$. Let $\angle{BCD}=\gamma$. By the Law of Cosines and keeping in mind that $\alpha$ and $\gamma$ are supplementary, we have

$$a^2+d^2-2ad\cos{\alpha}=b^2+c^2-2bc\cos{(180^\circ-\alpha)}$$

Yielding $\cos{\alpha}=\frac{a^2+d^2-b^2-c^2}{2(ad+bc)}$. Now, making use of the half angle formula for cosine,

$$\begin{aligned} \cos^2{\frac{\alpha}{2}}&=\frac{2ad+2bc+a^2+d^2-b^2-c^2}{4(ad+bc)}\\ &=\frac{(a+d)^2-(b-c)^2}{4(ad+bc)}\\&=\frac{(a+b-c+d)(a-b+c+d)}{4(ad+bc)}\\&=\frac{1}{ad+bc}\left(\frac{a+b+c+d}{2}-c\right)\left(\frac{a+b+c+d}{2}-b\right)\\&=\frac{(s-b)(s-c)}{ad+bc}\end{aligned}$$

$\square$

A similar procedure gives us the formula $\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad+bc}$.

Remark. Notice that we can get the analogous formulae for a triangle by making $b$ or $c$ equal zero. 

No explicit formulae are given in Durell, C. V.; Robson, A. (2003) [1930], Advanced Trigonometry for $\sin^2{\frac{\alpha}{2}}$ and $\cos^2{\frac{\alpha}{2}}$ in terms of the semiperimeter and sides lengths of a cyclic quadrilateral. Wikipedia neither provides such formulae.

A Proof of the Brahmagupta's formula
Denote $\Delta_c$ the area of the cyclic quadrilateral, $ABCD$. Then

$$\begin{aligned}\Delta_c&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{(180^\circ-\alpha)}}{2}\\&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\alpha}}{2}\\&=\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}(ad+bc)\\&=\sqrt{\frac{(s-a)(s-d)}{ad+bc}}\sqrt{\frac{(s-b)(s-c)}{ad+bc}}(ad+bc)\\&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\end{aligned}$$

$\square$

sábado, 13 de junio de 2020

Proof Without Words for the Addition Formula for Sine and Subtraction Formula for Cosine


I just hope this is new. What better gift than that on my birthday? :)

$$\sin{(\alpha+\beta)}=\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}\qquad\left(\alpha+\beta<\frac{\pi}{2}\right)$$

$$\cos{(\alpha-\beta)}=\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}\qquad\left(\alpha-\beta<\frac{\pi}{2}\right)$$



Related material.

Geometric Proof of the Sum Angle Formula (Sine, Cosine)


I wouldn't say the simpler one but at least $\sin(α+β)$ can be obtained almost without words.



$\frac{\cos{\alpha}}{\cos{\beta}}-\frac{\sin{\beta}(\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta})}{\cos{\beta}}=\frac{\cos{\alpha}-\sin^2{\beta}\cos{\alpha}-\sin{\beta}\sin{\alpha}\cos{\beta}}{\cos{\beta}}=\frac{\cos{\alpha}(1-\sin^2{\beta})-\sin{\beta}\sin{\alpha}\cos{\beta}}{\cos{\beta}}=$

$\frac{\cos{\alpha}\cos^2{\beta}-\sin{\beta}\sin{\alpha}\cos{\beta}}{\cos{\beta}}=\cos{\alpha}\cos{\beta}-\sin{\beta}\sin{\alpha}=\cos{(\alpha+\beta)}.$

viernes, 5 de junio de 2020

Yet Another Proof of the Law of Cosines

The law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. Using standard notation, the law of cosines states

$$c^2=a^2+b^2-2ab\cos{\gamma},$$

where $\gamma$ denotes the angle contained between sides of lengths $a$ and $b$ and opposite the side of length $c$. For the same figure, the other two relations are analogous:

$$a^2=b^2+c^2-2ac\cos{\alpha},$$
$$b^2=a^2+c^2-2ac\cos{\beta}.$$

Proof. Let $D$, $E$ and $F$ be the contact points of the incircle with $AC$, $AB$ and $BC$, respectively. Also, let $AE=AD=x$; $BE=BF=y$; $CD=CF=z$. We start from two well-known relationships of a triangle: $$\sin^2{\frac{\gamma}{2}}=\frac{(s-a)(s-b)}{ab} \qquad\text{and}\qquad \cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}$$  
(See Cut-the-knot's Relations between various elements of a triangle for proofs), where $s$ denotes the semiperimeter of $\triangle{ABC}$. Since $(s-a)=x$, $(s-b)=y$ and $(s-c)=z$, then the following identity holds:
$$ab\cos{\gamma}=ab\cos^2{\frac{\gamma}{2}}-ab\sin^2{\frac{\gamma}{2}}=sz-xy$$
Substituting and multiplying by 4, 
$$4ab\cos{\gamma}=(a+b+c)(a+b-c)-(b+c-a)(a+c-b)$$
Simplifying,
$$2ab\cos{\gamma}=a^2+b^2-c^2$$
$\square$

A similar reasoning must show that $a^2=b^2+c^2-2bc\cos{\alpha}$ and $b^2=a^2+c^2-2ac\cos{\beta}$.

AcknowledgementMy sincerest thanks to Angina Seng for giving helpful comments which allowed me to simplify the proof.

Related material.

miércoles, 20 de mayo de 2020

Another Simple Proof of Johnson's Theorem

Other proofs can be found in cut-the-knot.org. See also Johnson's Three Circles Theorem Revisited and Johnson's theorem proof.

Johnson's theorem: Let three equal circles with centers $J_a$, $J_b$, and $J_c$ intersect in a single point $H$ and intersect pairwise in the points $A$, $B$, and $C$. Then the circumcircle of the triangle $\triangle{ABC}$ is congruent to the original three.


Proof. Since $\odot{ABH}$ and $\odot{BCH}$ are congruent and as $\angle{BAH}$ and $\angle{BCH}$ are angles subtended by the same arc, it follows that $\angle{BAH}=\angle{BCH}$. Analogously, $\angle{CAH}=\angle{CBH}$. Let $O$, $J_a$ be the centers of $\odot{ABC}$ and $\odot{BCH}$, respectively, then

$$\angle{COB}=2\angle{CAB}=2\angle{CAH}+2\angle{BAH}=2\angle{CBH}+2\angle{BCH}=\angle{BJ_aC}.$$

It follows that $\triangle{BCO}$ and $\triangle{BCJ_a}$ are similar isosceles triangles sharing a common side, $BC$, so by $ASA$ we deduce that $\triangle{BCO}\cong{\triangle{BCJ_a}}$. Consequently, $\odot{ABC}$ is congruent to the original three. 

$\square$

Note: The point $H$ may cross the side lines of the triangle $\triangle{ABC}$ in points either interior or exterior to the sides. The reasoning in cases other than that considered above requires only minor adjustments.

sábado, 9 de mayo de 2020

The Pythagorean Theorem by Reductio ad Absurdum

Other proofs by contradiction can be found in cut-the-knot.org (see proof #122) and in Loomis' collection (see proofs # 16/2 and # 32).

Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Let $D$, $E$ and $F$ be the contact points of the incircle with $BC$, $AC$ and $AB$, respectively. Also, let $AE=AF=x$; $BD=BF=y$; $CD=CE=r$, where $r$ is the inradius of $\triangle{ABC}$.




Assume to the contrary that $a^2+b^2>c^2$. Then, 

$$(r+y)^2+(x+r)^2>(x+y)^2.$$
Expanding, collecting like terms and simplifying we get
$$ry+rx+r^2>xy.$$

Notice that $ry+rx+r^2=r(y+x+r)=rs=\Delta$, where $\Delta$ denotes the area of $\triangle{ABC}$ and $s$ its semiperimeter. Moreover, it is well-known that $xy=\Delta$. So $ry+rx+r^2>xy$ is equivalent to write $\Delta>\Delta$, which is a contradiction. A similar situation arise if you assume $a^2+b^2<c^2$.

There is a simple direct proof starting from $\Delta=rs$ and $\Delta=xy$, can you find it? My sincerest thanks to Andrius Navas and José Hernández for pointing me out this. 

jueves, 7 de mayo de 2020

Yet Another Proof of the Pythagorean Theorem - Simpler version

In cut-the-knot.org you can find more than a hundred proofs of the Pythagorean theorem. Here I give another proof which I hope to be new.

Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Below I will be using standard notations: $a$, $b$, $c$, for the side length, $s$ for the semiperimeter, $\Delta$ for the area and $r$ for the inradius.

$$s=c+r$$
$$s-\frac{c}{2}=\frac{c}{2}+r$$
$$\frac{a}{2}+\frac{b}{2}=\frac{c}{2}+r$$
$$a+b=c+2r$$

Squaring both sides, 

$$a^2+2ab+b^2=c^2+4cr+4r^2$$

It is well-known $\Delta=rs$, which implies $2ab=4r(c+r)$. So the equation $a^2+2ab+b^2=c^2+4cr+4r^2$ can be rewritten like this $a^2+b^2=c^2$, which is the Pythagorean theorem.

Below is the comment by John Molokach, whom I consider an expert on this topic (he has proved the theorem in more than ten different ways).