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 = 2s-2c = 2(s-c) and similarly for a and b. We have from the cosine theorem Sin(A/2) 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) So . . Again, by symmetry there are similar expressions involving the angles B and C. Sin(A) and Heron's formula 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.