Circles and Sines and Cosines
D. Atkinson
1
Area of a Circle – Archimedes
Let C be the area of a circle of radius r, and let T be the area of a right
triangle of height r and base c, where c is the circumference of the circle (see
Figure 1). Archimedes proves that C = T . He does this by showing
(a) If C < T then C > T , which is a contradiction.
(b) If C > T then C < T , which is a contradiction.
Since C < T or C = T or C > T are exhaustive, and since the first and third
options are contradictory, C = T must be true.
r -
r
c
Figure 1: Archimedes’ proof
Part (a) of the proof proceeds as follows: suppose C < T , so T − C > 0. Let
Pn be the area of an n-sided regular polygon that is inscribed in the circle
(see Figure 2, in which I have drawn an octogon as an example). For all n,
Pn is clearly less than C, but C − Pn decreases monotonically to zero as n
increases to infinity. Therefore we can find an N such that C − PN is less
than T − C:
C − PN < T − C
so
PN > 2C − T .
(1)
1
However, the area of the inscribed N -sided regular polygon is N times the
area of a triangle of base p and height h (see Figure 2), where N p, the
circumference of the inscribed polygon, is less than c (the circumference of
the circle), and h is less than r (the radius of the circle). So
PN = 21 N ph < 21 cr = T .
(2)
From Eqs. (1) and (2) we have
T > PN > 2C − T
so
T >C,
which contradicts the hypothesis, C < T , thus ruling it out as a possibility.
r
6
h
p
r
?
Figure 2: Inscribed octogon
Part (b) of the proof proceeds analogously: suppose C > T , so C − T > 0.
Let Pn0 be the area of an n-sided regular polygon that is circumscribed about
the circle (see Figure 3, again with an octogon). For all n, Pn0 is clearly
greater than C, but Pn0 − C decreases monotonically to zero as n increases to
infinity. Therefore we can find an N 0 such that PN0 0 − C is less than C − T :
PN0 0 − C < C − T
so
PN0 0 < 2C − T .
(3)
However, the area of the N 0 -sided circumscribed regular polygon is N 0 times
the area of a triangle of base p0 and height r (see Figure 3), where N p0 , the
circumference of the circumscribed polygon, is greater than c (the circumference of the circle). So
PN0 0 = 12 N 0 pr > 12 cr = T .
2
(4)
From Eqs. (3) and (4) we have
T < PN0 0 < 2C − T
so
T <C,
which contradicts the hypothesis, C > T , thus ruling it out as a possibility.
6
r
p0
?
Figure 3: Circumcribed octogon
This concludes Archimedes’ proof that C = T , namely 21 cr.
2
Definition of π
The transcendental number π can be defined as the ratio of the circumference,
c, of a circle to its diameter, d, and since the diameter is twice the radius, r,
it follows that π = 2rc , or equivalently,
c = 2πr .
(5)
The proof of Archimedes tells us that the area of the circle is
C = 21 cr =
1
2
× 2πr × r = πr2 .
The natural unit of angle (one radian) is defined to be the angle subtended
at the centre of a circle by an arc of length equal to the radius of the circle.
Therefore from Eq.(5) it follows that the angle subtended by the whole circle
is 2π, and the angle subtended by an an arc of length ε is ε. The area
ε
of a segment of angle ε is 2π
times the area of the whole circle, that is
ε
2
2
1
× πr = 2 εr .
2π
3
3
Sines and Cosines
We shall first prove two identities involving sines and cosines.
B
C
x
1
D
y
x
A
F
E
Figure 4: Addition formulae for sines and cosines
With AB = 1 to set the scale, we see from Figure 4 that
BD
DE
CD
AE
BC
BF
=
=
=
=
=
=
sin y
AD = cos y
AD sin x = sin x cos y
BD cos x = cos x sin y
AD cos x = cos x cos y
BD sin x = sin x sin y
sin(x + y)
AF = cos(x + y)
Since BF = CE = CD + DE we obtain by substitution
sin(x + y) = sin x cos y + cos x sin y ;
(6)
and from AF = AE − F E = AE − BC
cos(x + y) = cos x cos y − sin x sin y .
In the special case x =
π
2
(7)
− y, Eq.(6) yields sin π2 = cos2 y + sin2 y, that is
1 = sin2 y + cos2 y ;
4
(8)
and in the special case x = y, Eq.(7) yields
cos 2y = cos2 y − sin2 y .
(9)
On subtracting Eq.(9) from Eq.(8), we obtain
1 − cos 2y = 2 sin2 y .
4
(10)
Derivative of Sine and Cosine
To calculate the derivative of sin x, we use Eq.(6) and Eq.(10) to evaluate
1 − cos ε
sin ε
sin(x + ε) − sin x
= − sin x
+ cos x
ε
ε
ε
sin ε
sin2 (ε/2)
+ cos x
= −2 sin x
ε
ε
!2
sin(ε/2)
sin ε
= − 12 sin x
ε + cos x
,
ε/2
ε
(11)
and we are interested in the limit of this expression as ε tends to zero.
E
D
C
ε
A
1
B
Figure 5: Limiting behaviour
In Figure 5, two right triangles, ACB and ABE, are depicted, as well as the
segment BD of a circle of angle ε. If we specify the line AB to be of unit
5
length, then we see from triangle ABC that BC = sin ε and AC = cos ε. In
the limit that ε tends to zero, it is clear that BC tends to zero and AC tends
to one, since it becomes indistinguishable from AB. Thus
lim sin ε = 0
ε→0
and
lim cos ε = 1 .
(12)
ε→0
From Figure 5 we see that
area of triangle ABC < area of segment ABD < area of triangle ABE .
The area of triangle ABC is 21 × BC × AC = 12 sin ε cos ε, whilst the area of
triangle ABE is 21 × AB × BE = 21 tan ε. The area of the segment ABD is
2
1
2 εr , with r = AB = 1. So we have
1
2
Divide throughout by
1
2
sin ε cos ε < 12 ε < 21 tan ε .
sin ε:
cos ε <
ε
< sec ε ;
sin ε
sec ε >
sin ε
> cos ε .
ε
and now invert everything:
As ε tends to zero, we see from Eq.(12) that cos ε tends to unity. In fact,
cos ε tends to unity from below, while its inverse, sec ε, tends to unity from
above. Since sinε ε is squeezed in the middle, it must be so that
sin ε
−→ 1
ε
as
ε → 0.
So the factors ( sin(ε/2)
) and ( sinε ε ) in Eq.(11) tend to unity as ε tends to zero,
ε/2
and we are left with
d
sin(x + ε) − sin x
def
sin x = lim
= cos x .
ε→0
dx
ε
Putting x =
π
2
− y, we have sin x = cos y, cos x = sin y and
d
cos y = − sin y .
dy
6
d
dx
d
= − dy
, so