Geometry with Trigonometry

Sec. 9.4 Half angles 139

(i)

cos90F= 0 ,sin90F= 1 ,cos180F=− 1 ,

sin180F= 0 ,cos270F= 0 ,sin270F=− 1.

(ii) 2( (^90) F)= (^180) F, 2 ( (^180) F)= (^0) Fso that− (^180) F= (^180) F, and (^90) F+ (^270) F=
(^0) Fso that (^270) F=− (^90) F.
(iii) For allα∈A(F),
cos(α+ (^90) F)=−sinα,sin(α+ (^90) F)=cosα,
cos(α+ (^180) F)=−cosα,sin(α+ (^180) F)=−sinα,
cos(α+ (^270) F)=sinα,sin(α+ (^270) F)=−cosα.
Proof.
(i) These follow immediately from 9.2.1.
(ii) These follow from 9.2.1 and 9.3.4.
(iii) These follow immediately from 9.3.3 and (i) of the present theorem.

9.4 Halfangles...............................

9.4.1 .....................................

Definition. Given any angleα∈A(F)with support|QOP, its indicatori(α)meets
C(O;k)in a unique pointP′which is inH 1. Then the wedge or straight angle in
A(F)with support|QOP′is denoted by^12 αand is called ahalf-angle.
Given any angleα∈A(F), the equation 2 γ=αhas exactly two solutions in

A(F), these being^12 αand^12 α+ (^180) F.
Proof. In the definition we have
1
2 α=∠FQOP
′and takeP′′so
thatO =mp(P′,P′′).Letβ=
∠FQOP′′ so that β =^12 α+
(^180) F.ThensOP′(Q)=Pso that
by 9.3.3, 2(^12 α)=α, 2 β=α.
Now suppose thatγ,δ∈A(F)
and 2γ= 2 δ=α.Then cos2δ=
cos2γ so that 2cos^2 δ− 1 =
2cos^2 γ− 1 ,and hence cosδ=
±cosγ.

P

P′

O

J

I Q

R

S

T

i(α)

α

1

2 α

H 1

H 2

H 4 H 3

Figure 9.11.

Then also sin^2 δ=sin^2 γ so sinδ =±sinγ. Moreover sin2δ=sin2γ so
2sinδcosδ=2sinγcosγ.

We first suppose thatα= (^180) Fso that cosα=−1 and so cosγ=0. Then if
cosδ=cosγwe must have sinδ=sinγ,andsoδ=γ. Alternatively we must have
cosδ=−cosγ,sinδ=−sinγand soδ=γ+ (^180) F.