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
cos
(
A
)
=
b
2
+
c
2
−
a
2
2
b
c
{\displaystyle \displaystyle \cos(A)={{b^{2}+c^{2}-a^{2}} \over {2bc}}}
2
sin
2
(
A
2
)
=
1
−
cos
(
A
)
=
1
−
b
2
+
c
2
−
a
2
2
b
c
=
(
a
+
b
−
c
)
(
a
−
b
+
c
)
2
b
c
=
2
(
s
−
b
)
(
s
−
c
)
b
c
{\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
sin
(
A
2
)
=
(
s
−
b
)
(
s
−
c
)
b
c
{\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.
edit
2
cos
2
(
A
2
)
=
1
+
cos
(
A
)
=
1
+
b
2
+
c
2
−
a
2
2
b
c
=
(
a
+
b
+
c
)
(
b
+
c
−
a
)
2
b
c
=
2
s
(
s
−
a
)
b
c
{\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
cos
(
A
2
)
=
s
(
s
−
a
)
b
c
{\displaystyle \displaystyle \cos \left({\frac {A}{2}}\right)={\sqrt {{s(s-a)} \over {bc}}}}
tan
(
A
2
)
=
sin
(
A
2
)
cos
(
A
2
)
=
(
s
−
b
)
(
s
−
c
)
s
(
s
−
a
)
{\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.
A formula for sin(A) can be found using either of the following identities:
sin
(
A
)
=
2
sin
(
A
2
)
cos
(
A
2
)
{\displaystyle \displaystyle \sin(A)=2\sin \left({\frac {A}{2}}\right)\cos \left({\frac {A}{2}}\right)}
sin
(
A
)
=
(
1
+
cos
(
A
)
)
(
1
−
cos
(
A
)
)
{\displaystyle \displaystyle \sin(A)={\sqrt {(1+\cos(A))(1-\cos(A))}}}
These both lead to
sin
(
A
)
=
2
b
c
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
{\displaystyle \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
Δ
=
1
2
b
c
sin
(
A
)
{\displaystyle \displaystyle \Delta ={\frac {1}{2}}bc\sin(A)}
Δ
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
{\displaystyle \displaystyle \Delta ={\sqrt {s(s-a)(s-b)(s-c)}}}
which is Heron's formula.
