Half-Angle formula

Write $s=\dfrac{a+b+c}2$ for the semi-perimeter of $\triangle ABC$. The half-angles of the triangle can be expressed purely in terms of the three sides and $s$.

Theorem 3.6 (Half-angle formulas). In $\triangle ABC$,

Proof (of the $\sin\frac A2$ form). Start from $\sin^2\frac A2=\dfrac{1-\cos A}2$ and substitute the Law of Cosines value $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$:

Now $a-b+c=(a+b+c)-2b=2s-2b=2(s-b)$ and $a+b-c=2s-2c=2(s-c)$, so

(the positive square root, since $A/2\in(0^\circ,90^\circ)$ makes $\sin\frac A2>0$). The $\cos\frac A2$ and $\tan\frac A2$ forms follow the same way (or from $\tan\frac A2=\sin\frac A2/\cos\frac A2$). $\blacksquare$

By the identical argument (cycling $A\to B\to C\to A$, $a\to b\to c\to a$), the companion formulas for $B$ and $C$ are

Corollary. Using $\sin A=2\sin\frac A2\cos\frac A2$,

Heron's Formula. Named after Hero (Heron) of Alexandria, the Greek engineer-mathematician of the 1st century CE, this formula gives the area directly from the three sides alone — no angle needs to be found first.

Theorem 3.7. $\triangle=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\dfrac{a+b+c}2$.

Proof. From the area formula $\triangle=\tfrac12ab\sin C$ and $\sin C=2\sin\frac C2\cos\frac C2$,

Remarks.

1. Heron's formula and Pythagoras' theorem are interderivable for a right triangle — each can be used to establish the other.
2. If the area of a triangle must come out an integer, Heron's formula is the natural tool for hunting triangles with integer sides and integer area.
3. For a fixed perimeter, Heron's formula is useful for finding triangles with integer sides and integer area — e.g. a triangle of perimeter $100$ m with sides $32,34,34$ m has area exactly $480$ m².
   Note

   Worked illustrations from the textbook (own-words summary), Examples 3.56–3.63:

   • A circular-park segment problem: a chord of length $4$ km in a circle of diameter $8$ km. The Law of Cosines pins down the central angle as $\theta=\pi/3$, and the segment-area formula then gives the area set aside for the facility as $\tfrac43(2\pi-3\sqrt3)$ km².
   • A sine-rule identity, $b^2\sin2C+c^2\sin2B=2bc\sin A$, proved by writing $a,b,c$ as $2R\sin A,2R\sin B,2R\sin C$ and simplifying with $\sin2\theta=2\sin\theta\cos\theta$ and $A+B+C=\pi$.
   • A half-angle identity, $\sin\!\left(\tfrac{B-C}2\right)=\tfrac{b-c}a\cos\tfrac A2$, again via the sine rule and the sum-to-product formulas.
   • An angle-ratio problem: if the three angles of a triangle are in the ratio $1:2:3$ (i.e. $30^\circ,60^\circ,90^\circ$), the sine rule shows the sides come out in the ratio $1:\sqrt3:2$.
   • A projection-formula identity, $(b+c)\cos A+(c+a)\cos B+(a+b)\cos C=a+b+c$, proved by regrouping the left side into three projection-formula triples.
   • A sine-rule identity, $\dfrac{a^2+b^2}{a^2+c^2}=\dfrac{1+\cos(A-B)\cos C}{1+\cos(A-C)\cos B}$, via $a=2R\sin A$ etc. together with the product-to-sum identities.
   • An alternative derivation of the cosine rule directly from the sine rule — a self-check that the two laws are consistent with one another.
   • The isoperimetric result for triangles: using the AM–GM inequality on $(s-a),(s-b),(s-c)$ inside Heron's formula shows that, for a fixed perimeter, the area of a triangle is maximised exactly when $a=b=c$ — the equilateral triangle beats every other triangle of the same perimeter for area, with maximum area $\dfrac{s^2}{3}\sqrt3$ (attained at $a=b=c=\tfrac{2s}3$). This exact result is what Exercise 3.9, Q9 and Q10 (below) apply. …