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.