Half-Angle Formulas

10.5 Half-Angle Formulas

Theorem. With $s=\dfrac{a+b+c}{2}$,

and cyclically for $\frac B2,\frac C2$.

Proof. Start from the cosine rule, $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, and the double-angle identity $\cos A=1-2\sin^2\frac A2$. Then

Since $s=\frac{a+b+c}2$, we have $a+c-b=2(s-b)$ and $a+b-c=2(s-c)$. Substituting,

Similarly, using $\cos A=2\cos^2\frac A2-1$,

so $\cos^2\frac A2=\dfrac{s(s-a)}{bc}$. Dividing the two, $\tan^2\frac A2=\dfrac{(s-b)(s-c)}{s(s-a)}$. Taking positive square roots (each half-angle is acute, since $0<A<180^\circ$) gives all three formulas. $\blacksquare$

Worked example. For $a=6,b=8,c=10$ (a $6$–$8$–$10$ right triangle): $s=12$, so $\sin\frac A2=\sqrt{\frac{(12-8)(12-10)}{8\cdot10}}=\sqrt{\frac{8}{80}}=\sqrt{0.1}\approx0.316$, giving $\frac A2\approx18.43^\circ$, i.e. $A\approx36.87^\circ$ — matching the known angle of a $3$–$4$–$5$-type right triangle.

These formulas are the bridge to Heron's formula (§10.6) and to every inradius/exradius theorem in §§10.7–10.8, all of which are proved by substituting a half-angle expression into $r=(s-a)\tan\frac A2$ or its exradius analogue.

Connection to the incircle. Because $s-a,s-b,s-c$ are precisely the tangent lengths from $A,B,C$ to the incircle (§10.7), the half-angle formulas can be read geometrically: $\tan\frac A2=\dfrac{r}{s-a}$ combines the incircle's radius $r$ with the tangent length $s-a$ — so this section is not an isolated algebraic waypoint but the direct algebraic ancestor of every incircle and excircle theorem that follows.

A second worked example. For the same triangle $a=6,b=8,c=10,s=12$: $\cos\frac A2=\sqrt{\dfrac{s(s-a)}{bc}}=\sqrt{\dfrac{12\times6}{80}}=\sqrt{0.9}\approx0.949$, consistent with $\cos18.43^\circ\approx0.949$ found from $A\approx36.87^\circ$. As a further check, $\tan\frac A2=\dfrac{\sin\frac A2}{\cos\frac A2}\approx\dfrac{0.316}{0.949}\approx0.333=\dfrac13$, and indeed $\tan18.43^\circ\approx0.333$. …