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 = Template:Frac(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

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

Sin(A/2)

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

So

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

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)

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

So

$\cos \left(\frac{A}{2}\right)=\sqrt{\frac{s(s-a)}{bc}}$.

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

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:

$\sin (A)=2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)$

$\sin (A)=\sqrt{(1+\cos (A))(1-\cos (A))}$

These both lead to

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

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

$\mathrm{\Delta }=\frac{1}{2}bc\sin (A)$,

$\mathrm{\Delta }=\sqrt{s(s-a)(s-b)(s-c)}$

which is Heron's formula.