Cosine of Half Angle in Triangle

Theorem
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Let $s$ denote the semiperimeter of $\triangle ABC$.

Then:
 * $\cos \dfrac C 2 = \sqrt {\dfrac {s \paren {s - c} } {a b} }$

Also see

 * Sine of Half Angle in Triangle
 * Tangent of Half Angle in Triangle