Page 17 - MATINF Nr. 13-14

A Generalization of the Law of Cotangents 17

Lemma (Half–angle formulas). For α = ∠BAD [1, p. 186],

α (s − a)(s − d) α (s − b)(s − c)
sin 2 = , cos 2 = .
2 ad + bc 2 ad + bc

Proof. Let γ = ∠BCD. Since ABCD is cyclic, α + γ = π. Applying the Law of Cosines in
triangles ABD and BCD and using cos(π − α) = − cos α,

2
2
2
2
2
2
a + d − 2ad cos α = b + c − 2bc cos(π − α) = b + c + 2bc cos α,
hence
2
2
2
a + d − b − c 2
cos α = .
2(ad + bc)
2
For cos (α/2):
2
2
2
α 1 + cos α 2(ad + bc) + a + d − b − c 2
cos 2 = =
2 2 4(ad + bc)
2
(a + d) − (b − c) 2 (a + d − b + c)(a + d + b − c)
= =
4(ad + bc) 4(ad + bc)

(a + b + c + d) − 2b (a + b + c + d) − 2c
=
4(ad + bc)
(s − b)(s − c)
= .
ad + bc
2
For sin (α/2):
2
2
2
2
α 1 − cos α 2(ad + bc) − (a + d − b − c )
sin 2 = =
2 2 4(ad + bc)
2
(b + c) − (a − d) 2 (b + c − a + d)(b + c + a − d)
= =
4(ad + bc) 4(ad + bc)

(a + b + c + d) − 2a (a + b + c + d) − 2d
=
4(ad + bc)
(s − a)(s − d)
= .
ad + bc
This proves the two identities for α.

Theorem (Generalized law of cotangents for cyclic quadrilaterals). As a consequ-
ence of the half–angle formulas,

cot(α/2) cot(β/2) cot(γ/2) cot(δ/2) 1
= = = = .
(s − b)(s − c) (s − c)(s − d) (s − d)(s − a) (s − a)(s − b) ∆

Proof. From the lemma for α,
2
α cos (α/2) (s − b)(s − c)
cot 2 = = ,
2
2 sin (α/2) (s − a)(s − d)