# Trigonometry/Solving triangles by half-angle formulae

In this section, we present alternative ways of solving triangles by using half-angle formulae.

Given a triangle with sides a, b and c, define

s = 12(a+b+c).

Note that

a+b-c = 2s-2c = 2(s-c)

and similarly for a and b.

We have from the cosine theorem

${\displaystyle \displaystyle \cos(A)={{b^{2}+c^{2}-a^{2}} \over {2bc}}}$

## Sin(A/2)Edit

${\displaystyle \displaystyle 2\sin ^{2}\left({\frac {A}{2}}\right)=1-\cos(A)=1-{{b^{2}+c^{2}-a^{2}} \over {2bc}}={{(a+b-c)(a-b+c)} \over {2bc}}={{2(s-b)(s-c)} \over {bc}}}$

So

${\displaystyle \displaystyle \sin \left({\frac {A}{2}}\right)={\sqrt {{(s-b)(s-c)} \over {bc}}}}$ .

By symmetry, there are similar expressions involving the angles B and C.

Note that in this expression and all the others for half angles, the positive square root is always taken. This is because a half-angle of a triangle must always be less than a right angle.

## Cos(A/2) and tan(A/2)Edit

${\displaystyle \displaystyle 2\cos ^{2}\left({\frac {A}{2}}\right)=1+\cos(A)=1+{{b^{2}+c^{2}-a^{2}} \over {2bc}}={{(a+b+c)(b+c-a)} \over {2bc}}={{2s(s-a)} \over {bc}}}$

So

${\displaystyle \displaystyle \cos \left({\frac {A}{2}}\right)={\sqrt {{s(s-a)} \over {bc}}}}$ .
${\displaystyle \displaystyle \tan \left({\frac {A}{2}}\right)={\sin({\frac {A}{2}}) \over \cos({\frac {A}{2}})}={\sqrt {{(s-b)(s-c)} \over {s(s-a)}}}}$ .

Again, by symmetry there are similar expressions involving the angles B and C.

## Sin(A) and Heron's formulaEdit

A formula for sin(A) can be found using either of the following identities:

${\displaystyle \displaystyle \sin(A)=2\sin \left({\frac {A}{2}}\right)\cos \left({\frac {A}{2}}\right)}$
${\displaystyle \displaystyle \sin(A)={\sqrt {(1+\cos(A))(1-\cos(A))}}}$

${\displaystyle \displaystyle \sin(A)={\frac {2}{bc}}{\sqrt {s(s-a)(s-b)(s-c)}}}$
${\displaystyle \displaystyle \Delta ={\frac {1}{2}}bc\sin(A)}$ ,
${\displaystyle \displaystyle \Delta ={\sqrt {s(s-a)(s-b)(s-c)}}}$