In Cut-the-knot's Relations between various elements of a triangle, the formulae (a generalization can be found here)
$$\sin^2{\frac{\gamma}{2}} = \frac{(s-a)(s-b)}{ab}\quad and\quad\cos^2{\frac{\gamma}{2}}= \frac{s(s-c)}{ab}$$
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,
$$\begin{aligned} 1-\sin^2{\frac{\gamma}{2}} &= \frac{s(s-c)}{ab} \\ \sin^2{\frac{\gamma}{2}} &= 1-\frac{s(s-c)}{ab} \\ &= \frac{\left(ab-s(s-c)\right)}{ab} \\ &= \frac{\left((y+z)(x+z)-(x+y+z)(z)\right)}{ab}\\ &= \frac{xy}{ab} \\ &=\frac{(s-a)(s-b)}{ab} \end{aligned}$$
$\square$
Geometrical proofs can be found at Trigonography.com.
1. 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
$$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$
$\square$
2. 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,
$$4ab\cos{\gamma}=(a+b+c)(a+b-c)-(b+c-a)(a+c-b)$$
Expanding and 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 $b^2=a^2+c^2-2ac\cos{\beta}$.
3. 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.
$\square$
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
$$\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\frac{s}{r}$$
To prove the above identity we will show that the right hand side equals $\frac{s}{r}$.
$\square$
We invite the reader to prove the following identity (possibly new) on their own.
$$\frac{\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}+\cot{\frac{\alpha}{2}}\cot{\frac{\gamma}{2}}+\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}}{(s-a)(s-b)+(s-a)(s-c)+(s-b)(s-c)}=\frac{1}{r^2}$$
4. The product $AI\cdot{BI}\cdot{CI}$
Consider a triangle $\triangle{ABC}$ and its Incenter, $I$. Denote $R$ and $r$ the circumradius and inradius, respectively. Also let $AI=k$; $BI=l$; $CI=m$. Then, the following identity holds
$$klm=4Rr^2$$
Proof. We make use of the half-angle formula,
$$\cos^2{\frac{\gamma}{2}}= \frac{s(s-c)}{ab}$$
Notice that $\cos{\frac{\gamma}{2}}=\frac{(s-c)}{m}$. Also, because of half-angle formula we have $\cos{\frac{\gamma}{2}}=\sqrt{\frac{s(s-a)}{ab}}$. Equating both expressions and solving for $m^2$,
$$m^2=\frac{ab(s-c)}{s}$$
Similarly you get $k^2=\frac{bc(s-a)}{s}$ and $l^2=\frac{ac(s-b)}{s}$. Hence,
$$(klm)^2=\frac{a^2b^2c^2(s-a)(s-b)(s-c)}{s^3}=\frac{a^2b^2c^2s(s-a)(s-b)(s-c)}{s^4}$$
Substituting from Heron's formula,
$$(klm)^2=\frac{a^2b^2c^2\Delta^2}{s^4}$$
Simplifying and using the well-known formulas $abc=4R\Delta$ and $\Delta=rs$ you get the desired result.
$$klm=\frac{abc\Delta}{s^2}=\frac{4R\Delta^2}{s^2}=4Rr^2$$
See also
- Length of angle bisector: yet another application of the half-angle formulas
- Using the half-angle formulas to derive Mahavira's identities
- Using the half-angle formula for cosine to derive Zelich's lemma on mixtilinear incircles
- Two Identities and their Consequences, pp. 5-11
- Generalization of Mollweide's formulas (rather Newton's)
- Generalization of the law of tangents
- The compound angle formulas from the half angle formulas
- $\sum_{cyc}\tan\frac\alpha2\tan\frac\beta2\geq4$
- Another proof of Euler inequality via the half-angle formulas
