Geometry with Trigonometry

Sec. 12.2 Circular functions 241

Without loss of generality we suppose that|α|◦≤|β|◦,sothatbytheabove,
|^12 α|◦≤|^12 β|◦.As|^12 α|◦+|^12 β|◦≤180, then|^12 α|◦≤ 90 ,|^12 β|◦≤180, so thatP 1 ′∈
H 1 ,P 2 ′∈H 1 and[O,P 1 ′⊂IR(|QOP′ 2 ). Let the angle inA∗(F)with degree mea-

sure^12 (|α|◦+|β|◦)have support|QONwhere|O,N|=k.Thenas^12 (|α|◦+|β|◦)≤
180 we haveN∈H 1.
Let the angle inA∗(F)with degree measure^14 (|α|◦+|β|◦)have support|QOW,
where|O,W|=k.ThenW∈H 1 and[O,P 1 ′⊂IR(|QOW)as^12 |α|◦≤^14 (|α|◦+|β|◦).
Thus

|∠P′ 1 OW|◦=|∠QOW|◦−|∠QOP′ 1 |◦=^14 (|α|◦+|β|◦)−^12 |α|◦=^14 (|β|◦−|α|◦).

As^14 (|α|◦+|β|◦)≤^12 |β|◦we have[O,W ⊂IR(|QOP′ 2 ).Then

|∠WOP′ 2 |◦=|∠QOP′ 2 |◦−|∠QOW|◦=^12 |β|◦−^14 (|α|◦+|β|◦)=^14 (|β|◦−|α|◦).

As|∠P′ 1 OW|◦=|∠WOP′ 2 |◦we haveOW=midlP 1 ′OP 2 ′.
By a similar argument

|∠QOW|◦=^14 (|α|◦+|β|◦)=|∠WON|◦,

soOW=midlQON.ThenN=sOW(Q)and so^12 α⊕^12 βhas support|QON.
When^12 (|α|◦+|β|◦)<180 this gives thatP 3 ′=Nand so^12 α⊕^12 β=^12 (α⊕β).
The remaining case is when|α|◦+|β|◦=360. Then^12 (|α|◦+|β|◦)=180 so^12 α⊕
1
2 β=^180 F, whileα⊕β=^360 Fso

1
2 (α⊕β)=^180 F.
COROLLARY 1.Ifα,β∈A∗(F)are such that|α|◦+|β|◦≤ 360 , then for all
integers n≥ 1 ,
1
2 n

(α⊕β)=

1

2 n

α⊕

1

2 n

β.

COROLLARY 2.Ifα,β∈A∗(F)are such that|α|◦+|β|◦≤ 360 ,then

|α⊕β|◦=|α|◦+|β|◦.

Proof. In the course of the proof we saw that, under the given conditions,

|^12 α⊕^12 β|◦=^12 (|α|◦+|β|◦),^12 α⊕^12 β=^12 (α+β).

From these|^12 (α⊕β)|◦=^12 (|α|◦+|β|◦),sothat|α⊕β|◦=|α|◦+|β|◦.

12.2 Circularfunctions ...........................

12.2.1 .....................................

Definition.Letx∈Rsatisfy 0≤x≤360. Then if forα∈A∗(F)we have|α|◦=x,
we writec(x)=cosα,s(x)=sinα.By9.6,candsdo not depend onF.