# 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*=^{1}⁄_{2}(*a*+*b*+*c*).

Note that:

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

and similarly for *a* and *b*.

We have from the cosine theorem

## Sin(A/2)Edit

So

- .

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

So

- .

- .

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:

These both lead to

The positive square root is always used, since A cannot exceed 180º. Again, by symmetry there are similar expressions involving the angles B and C. These expressions provide an alternative proof of the sine theorem.

Since the area of a triangle

- ,

which is Heron's formula.