Proof of the Law of Sines and the Law of Cosines
Law of Sines.
Consider a triangle ABC inscribed in a circle with center O and radius r.
A
b
c
O
r
C
r
D
a/2
a/2
B
From basic geometry, we know that A = ∠CAB = ∠DOB. From the right triangle DOB,
we deduce
a/2
a
sin A =
=⇒
= 2r.
r
sin A
By repeating the same argument for angles B and C, we obtain the Law of Sines.
a
b
c
=
=
= 2r
sin A
sin B
sin C
Observe that the proof is similar if the center O is not inside triangle ABC. Since the
outside angle ∠COB satisfies ∠COB = 2A, we have ∠DOB = 180◦ − A. Since
sin(180◦ − A) = sin A,
we have sin A =
a/2
a
=⇒
= 2r.
r
sin A
A
c
b
a/2
a/2
C
D
r
r
O
2A
Gilles Cazelais. Typeset with LATEX on June 12, 2006.
B
Law of Cosines.
Consider the following triangle ABC.
C
b
a
h
A
B
D
x
c−x
From the right triangle ADC, we deduce
x2 + h2 = b2
(1)
and
x
=⇒ x = b cos A.
b
From the right triangle BDC, we deduce
cos A =
(c − x)2 + h2 = a2 =⇒ a2 = c2 − 2cx + (x2 + h2 ).
(2)
(3)
Substituting equations (1) and (2) into (3), we get
a2 = c2 − 2c(b cos A) + b2
which is the Law of Cosines.
a2 = b2 + c2 − 2bc cos A
The proof is similar if angle A is obtuse.
C
a
h
b
D
A
c
B
x
From the right triangle ADC, we deduce
x2 + h2 = b2
and
cos(180◦ − A) = − cos A =
x
=⇒ x = −b cos A.
b
(4)
(5)
From the right triangle BDC, we deduce
(c + x)2 + h2 = a2 =⇒ a2 = c2 + 2cx + (x2 + h2 ).
Substituting equations (4) and (5) into (6) gives the Law of Cosines.
(6)