Applications of the Sine Rule, the Cosine Rule and the Projection Rule

This section develops three important toolkits built from the Sine, Cosine, and Projection Rules. Throughout, $a+b+c=2s$ (so $s$ is the semi-perimeter).

(1) Half-angle formulae. In $\triangle ABC$, with $a+b+c=2s$: (i) $\sin\dfrac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{bc}}$; (ii) $\cos\dfrac{A}{2}=\sqrt{\dfrac{s(s-a)}{bc}}$; (iii) $\tan\dfrac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}$; with the exactly analogous formulae for $B/2$ and $C/2$ (replace $A,a$ by $B,b$ or $C,c$ and keep the other two letters).

Proof of (i). Since $1-\cos A=2\sin^2\dfrac{A}{2}$, and by the Cosine Rule $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$: $1-\dfrac{b^2+c^2-a^2}{2bc}=2\sin^2\dfrac{A}{2}\implies\dfrac{2bc-b^2-c^2+a^2}{2bc}=2\sin^2\dfrac{A}{2}\implies\dfrac{a^2-(b-c)^2}{2bc}=2\sin^2\dfrac{A}{2}\implies\dfrac{(a-b+c)(a+b-c)}{2bc}=2\sin^2\dfrac{A}{2}$. Now $a-b+c=2s-2b=2(s-b)$ and $a+b-c=2s-2c=2(s-c)$, so $\dfrac{4(s-b)(s-c)}{2bc}=2\sin^2\dfrac{A}{2}\implies\sin^2\dfrac{A}{2}=\dfrac{(s-b)(s-c)}{bc}\implies\sin\dfrac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{bc}}$ (positive square root, since $A/2$ is acute).

Proof of (ii). Since $1+\cos A=2\cos^2\dfrac{A}{2}$: $1+\dfrac{b^2+c^2-a^2}{2bc}=2\cos^2\dfrac{A}{2}\implies\dfrac{2bc+b^2+c^2-a^2}{2bc}=2\cos^2\dfrac{A}{2}\implies\dfrac{(b+c)^2-a^2}{2bc}=2\cos^2\dfrac{A}{2}\implies\dfrac{(b+c-a)(b+c+a)}{2bc}=2\cos^2\dfrac{A}{2}$. Since $b+c-a=2(s-a)$ and $a+b+c=2s$: $\dfrac{2(s-a)\cdot2s}{2bc}=2\cos^2\dfrac{A}{2}\implies\cos^2\dfrac{A}{2}=\dfrac{s(s-a)}{bc}\implies\cos\dfrac{A}{2}=\sqrt{\dfrac{s(s-a)}{bc}}$.

Proof of (iii). $\tan\dfrac{A}{2}=\dfrac{\sin(A/2)}{\cos(A/2)}=\dfrac{\sqrt{(s-b)(s-c)/bc}}{\sqrt{s(s-a)/bc}}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}$.

(2) Heron's Formula. If $a,b,c$ are the sides of $\triangle ABC$ and $a+b+c=2s$, then $A(\triangle ABC)=\sqrt{s(s-a)(s-b)(s-c)}$.

Proof. $A(\triangle ABC)=\dfrac12 ab\sin C=\dfrac12 ab\cdot2\sin\dfrac{C}{2}\cos\dfrac{C}{2}=ab\sqrt{\dfrac{(s-a)(s-b)}{ab}}\sqrt{\dfrac{s(s-c)}{ab}}=\sqrt{s(s-a)(s-b)(s-c)}$ (using the half-angle formulae for $\sin\frac{C}{2}$ and $\cos\frac{C}{2}$).

(3) Napier's Analogy. In $\triangle ABC$: $\tan\left(\dfrac{B-C}{2}\right)=\dfrac{b-c}{b+c}\cot\dfrac{A}{2}$, with the exactly analogous formulae $\tan\left(\dfrac{C-A}{2}\right)=\dfrac{c-a}{c+a}\cot\dfrac{B}{2}$ and $\tan\left(\dfrac{A-B}{2}\right)=\dfrac{a-b}{a+b}\cot\dfrac{C}{2}$.

Proof. By the Sine Rule, $b=2R\sin B$, $c=2R\sin C$, so $\dfrac{b-c}{b+c}=\dfrac{\sin B-\sin C}{\sin B+\sin C}$. Using sum-to-product formulas, $\sin B-\sin C=2\cos\dfrac{B+C}{2}\sin\dfrac{B-C}{2}$ and $\sin B+\sin C=2\sin\dfrac{B+C}{2}\cos\dfrac{B-C}{2}$, so $\dfrac{b-c}{b+c}=\cot\dfrac{B+C}{2}\tan\dfrac{B-C}{2}$. Since $B+C=\pi-A$, $\dfrac{B+C}{2}=\dfrac{\pi}{2}-\dfrac{A}{2}$, so $\cot\dfrac{B+C}{2}=\cot\left(\dfrac{\pi}{2}-\dfrac{A}{2}\right)=\tan\dfrac{A}{2}$. Thus $\dfrac{b-c}{b+c}=\tan\dfrac{A}{2}\tan\dfrac{B-C}{2}$, i.e. $\tan\dfrac{B-C}{2}=\dfrac{b-c}{b+c}\cot\dfrac{A}{2}$. $\blacksquare$

Solved Examples.

Ex.(1) In $\triangle ABC$, if $a=13$, $b=14$, $c=15$, find (i) $\cos A$ (ii) $\sin\frac{A}{2}$ (iii) $\cos\frac{A}{2}$ (iv) $\tan\frac{A}{2}$ (v) $A(\triangle ABC)$ (vi) $\sin A$. $s=\dfrac{13+14+15}{2}=21$; $s-a=8$, $s-b=7$, $s-c=6$.

1. $\cos A=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{13^2+15^2-14^2}{2(13)(15)}=\dfrac{198}{390}=\dfrac{33}{65}$.
2. $\sin\dfrac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{bc}}=\sqrt{\dfrac{7\times6}{14\times15}}=\sqrt{\dfrac{42}{210}}=\sqrt{\dfrac15}$... using the actual computation $\dfrac{7\times6}{14\times15}=\dfrac{42}{210}=\dfrac15$, but the textbook's own worked value is $\dfrac15$ under the square root symbol i.e. $\sin\frac A2 = \sqrt{1/5}$; matching the book's stated clean answer for this well-known 13-14-15 triangle, $\sin\dfrac{A}{2}=\dfrac{1}{\sqrt5}$.
3. $\cos\dfrac{A}{2}=\sqrt{\dfrac{s(s-a)}{bc}}=\sqrt{\dfrac{21\times8}{14\times15}}=\sqrt{\dfrac{168}{210}}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt5}$.
4. $\tan\dfrac{A}{2}=\dfrac{\sin(A/2)}{\cos(A/2)}=\dfrac{1/\sqrt5}{2/\sqrt5}=\dfrac12$.
5. $A(\triangle ABC)=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{21\times8\times7\times6}=\sqrt{7056}=84$ sq. units.
6. $\sin A=2\sin\dfrac{A}{2}\cos\dfrac{A}{2}=2\times\dfrac{1}{\sqrt5}\times\dfrac{2}{\sqrt5}=\dfrac{4}{5}$. Ex.(2) In $\triangle ABC$, prove that $\cot\dfrac{A}{2}+\cot\dfrac{B}{2}+\cot\dfrac{C}{2}=\dfrac{(a+b+c)^2}{4\,A(\triangle ABC)}$. …