Geometry with Trigonometry

246 Trigonometric functions in calculus Ch. 12

Proof. Suppose that 0<x<360, andh>0 is so small thatx+h< 360 ,x−h>0.
Then

c(x+h)−c(x)

h

=

c(x)c(h)−s(x)s(h)−c(x)

h

=c(x)

c(h)− 1

h

−s(x)

s(h)

h

Butc(h)h−^1 =−^2 s(

h 2 ) 2

h →^0 (h→^0 +)and so

c(x+h)−c(x)

h

→−π

180

s(x)(h→ 0 +).

Also

c(x)=c(x−h+h)=c(x−h)c(h)+s(x−h)s(h),

s(x)=s(x−h+h)=c(x−h)s(h)+s(x−h)c(h).

Asc(h)^2 +s(h)^2 =1 we can solve these equations by Cramer’s rule to obtain

c(x−h)=c(x)c(h)+s(x)s(h),s(x−h)=s(x)c(h)−c(x)s(h).

From the first of these, by a similar argument to that above, it follows that

c(x−h)−c(x)

−h

→−

π

180

s(x)(h→ 0 +).

These results combined show thatc′(x)=− 180 πs(x).
The other result is proved similarly.

12.4 Parametric equations for a circle ...................

12.4.1Areaofadisk .............................

The diskD(O;k)={Z:|O,Z|≤k}has areaπk^2.
Proof.
We use a well-known expression for area as the value of a line integral; see e.g.
Apostol [1, Volume II, page 383].
Ifγis the path with equationx=kc(t),y=ks(t)( 0 ≤t≤ 360 ),thenγtraverses the
circle( C(O,k)once. Moreover ifO≡( 0 , 0 ),P≡(kc(t),ks(t)),T≡(kc′(t),ks′(t)) =

−π 180 ks(t),πkc 180 (t)

)

,wehavethat

δF(O,P,T)=

π

360

k^2 [c^2 (t)+s^2 (t)] =

π

360

k^2 > 0.

NowC(O;k)is the boundary ofD(O;k)and so the area ofD(O;k)is given by

1

2

∫

γ

xdy−ydx=^12

∫ 360

0

k^2 [c(t)s′(t)−s(t)c′(t)]dt

=^12

∫ 360

0

π

180

k^2 dt=πk^2.