Trigonometry/Solving triangles by half-angle formulae

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

Sin(A/2)[edit]

\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 \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 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 \cos\left(\frac{A}{2}\right) = \sqrt{{s(s-a)} \over {bc}}.
\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 formula[edit]

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

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

These both lead to

\displaystyle \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

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

which is Heron's formula.