Geometry with Trigonometry

Sec. 9.3 Angles in standard position 133

Proof.LetQ,Rbe the points whereC(O;k)meets[O,I and[O,J, respectively;
thenQandRhave Cartesian coordinates(k, 0 )and( 0 ,k), respectively. LetU,V
be the feet of the perpendiculars fromZto the linesOIandOJ, so that these have
Cartesian coordinates(x, 0 )and( 0 ,y)respectively. NowO≡( 0 , 0 )andZ≡(x,y)so
by the distance formula(x− 0 )^2 +(y− 0 )^2 =k^2. Thusx^2 +y^2 =k^2 ,sothatx^2 ≤k^2
and ask>0wehavex≤k. Then by the distance formula

|Q,U|=

√

(k−x)^2 +( 0 − 0 )^2 =k−x,

ask−x≥0, and similarly|R,V|=k−y. Hence

cosα=

k−|Q,U|

k

=

k−(k−x)

k

=

x

k

,

sinα=

k−|R,V|

k

=

k−(k−y)

k

=

y

k

Thusx=kcosα,y=ksinα.
We refer tokandαaspolar coordinatesof the pointZwith respect toF.

9.2.3 .....................................

With the notation of 9.2.1, letαbe an angle with support|IOP=|QOPand indicator
i(α)inH 1. Then we have the following properties:-

(i)For allα,cos^2 α+sin^2 α=1.

(ii)For P∈Q 1 ,cosα≥ 0 ,sinα≥ 0 ;forP∈Q 2 ,cosα≤ 0 ,sinα≥ 0 ;forP∈

Q 3 ,cosα≤ 0 ,sinα≤ 0 ;forP∈Q 4 ,cosα≥ 0 ,sinα≤ 0.

Proof.

(i) As in the proof in 9.2.1,

cosα=±

|O,U|

|O,P|

,sinα=±

|O,V|

|O,P|

Now whenO,U,P,Vare not collinear they are the vertices of a rectangle and so
|O,V|=|U,P|. Then by Pythagoras’ theorem

|O,U|^2 +|U,P|^2 =|O,P|^2 ,

and the result follows. It can be verified directly whenPis any ofQ,R,S,T.
(ii) This follows directly from details in the proof in 9.2.1.

9.3 Angles in standard position ......................

9.3.1 Angles in standard position ......................

COMMENT. The second use that we make of the concept of indicator of an angle is
to identify angles with respect to a frame of reference.