There is an amazingly important inequality which to most of us is fairly obvious from the
picture. Here is a picture which illustrates the conclusion of this corollary.
PICT
The corollary states that the length of the subtended arc shown in the picture is
longer than the vertical side of the triangle and smaller than the sum of the vertical side
with the segment having length 1 − cosθ.
Corollary 2.3.10Let 0 < radian measure of θ < π∕4. Then letting A be the arc onthe unit circle resulting from situating the angle with one side on the positive x axis andthe other side pointing up from the positive x axis,
(1 − cosθ)+ sinθ ≥ l(A ) ≥ sin θ (2.13)
(2.13)
While this may seem obvious and in fact you could easily convince yourself of its
truth with graphing circles and using string, to do this right, one must give a more
mathematically precise treatment of arc length on the circle. First note that for t ≥ 0,
there is a unique nonnegative integer n such that
t = 2πn+ l
where l ∈ [0,2π). Similarly if t < 0 there is a unique nonnegative integer n such
that
t = (− 2π)n + l
where again we can have l ∈ [0,2π). Then to get to the point p
(t)
, one simply starts at
(0,0)
and considers on the unit circle and moves in the counter clockwise direction a
distance of l using the description of length of a circular arc about to be presented. Thus
it suffices to consider the length of an arc on the unit circle or more generally an arc on a
circle of radius r.
To give a precise description of what is meant by the length of an arc, consider the
following picture.
PICT
In this picture, there are two circles, a big one having radius R and a little one having
radius r. The angle θ is situated in two different ways subtending the arcs A_{1} and A_{2} as
shown.
Letting A be an arc of a circle, like those shown in the above picture, A subset of
A,
{p0,⋅⋅⋅,pn}
is a partition of A if p_{0} is one endpoint, p_{n} is the other end
point, and the points are encountered in the indicated order as one moves in
the counter clockwise direction along the arc. To illustrate, see the following
picture.
PICT
Also, denote by P
(A )
the set of all such partitions. For P =
{p0,⋅⋅⋅,pn}
, denote
by
|pi − pi−1|
the distance between p_{i} and p_{i−1}. Then for P ∈P
(A)
, define
|P |
≡∑_{i=1}^{n}
|pi − pi−1|
. Thus
|P|
consists of the sum of the lengths of the little lines
joining successive points of P and appears to be an approximation to the length of the
circular arc A. By the Pythagorean theorem the length of any of the straight line
segments joining successive points in a partition is smaller than the sum of the two sides
of a right triangle having the given straight line segment as its hypotenuse. For example,
see the following picture.
PICT
The sum of the lengths of the straight line segments in the part of the picture found
in the right rectangle above is less than A + B and the sum of the lengths of the straight
line segments in the part of the picture found in the left rectangle above is less than
C + D and this would be so for any partition. Therefore, for any P ∈P
(A)
,
|P |
≤ M
where M is the perimeter of a rectangle containing the arc A. To be a little sloppy,
simply pick M to be the perimeter of a rectangle containing the whole circle of which A
is a part. The only purpose for doing this is to obtain the existence of an upper bound.
Therefore,
{|P | : P ∈ P(A )}
is a set of numbers which is bounded above by
M and by completeness of ℝ it is possible to define the length of A,l
(A )
, by
l
(A)
≡ sup
{|P| : P ∈ P (A)}
.
A fundamental observation is that if P,Q ∈P
(A )
and P ⊆ Q, then
|P|
≤
|Q |
. To see
this, add in one point at a time to P. This effect of adding in one point is illustrated in
the following picture.
PICT
Also, letting
{p0,⋅⋅⋅,pn}
be a partition of A, specify angles, θ_{i} as follows. The angle
θ_{i} is formed by the two lines, one from the center of the circle to p_{i} and the other line
from the center of the circle to p_{i−1}. Furthermore, a specification of these angles yields
the partition of A in the following way. Place the vertex of θ_{1} on the center of the circle,
letting one side lie on the line from the center of the circle to p_{0} and the other side
extended resulting in a point further along the arc in the counter clockwise direction.
When the angles, θ_{1},
⋅⋅⋅
,θ_{i−1} have produced points, p_{0},
⋅⋅⋅
,p_{i−1} on the arc, place the
vertex of θ_{i} on the center of the circle and let one side of θ_{i} coincide with the side of the
angle θ_{i−1} which is most counter clockwise, the other side of θ_{i} when extended, resulting
in a point further along the arc A in the counterclockwise direction as shown
below.
PICT
Now let ε > 0 be given and pick P_{1}∈P
(A )
1
such that
|P |
1
+ ε > l
(A )
1
. Then
determining the angles as just described, use these angles to produce a corresponding
partition of A_{2}, P_{2}. If
|P |
2
+ ε > l
(A )
2
, then stop. Otherwise, pick Q ∈P
(A )
2
such that
|Q |
+ ε > l
(A )
2
and let P_{2}^{′} = P_{2}∪Q. Then use the angles determined by P_{2}^{′} to obtain
P_{1}^{′}∈P
(A )
1
. Then
|P′|
1
+ ε > l
(A )
1
,
|P′|
2
+ ε > l
(A )
2
, and both P_{1}^{′} and
P_{2}^{′} determine the same sequence of angles. Using Problem 36 on Page 168
about the base angles of an isosceles triangle, the two triangles are similar and
so
|P′1|= R-
|P′2| r
Therefore
l(A2) < |P ′|+ ε = r-|P ′|+ ε ≤ r-l(A1 )+ ε.
2 R 1 R
Since ε is arbitrary, this shows Rl
(A2 )
≤ rl
(A1)
. But now reverse the argument and
write
l(A1 ) < |P ′1|+ ε = R-|P′2|+ε ≤ R-l(A2 )+ ε
r r
which implies, since ε is arbitrary that Rl
(A2 )
≥ rl
(A1 )
and this has proved the
following theorem.
Theorem 2.3.11Let θ be an angle which subtends two arcs, A_{R}on acircle of radius R and A_{r}on a circle of radius r. Then denoting by l
(A )
the lengthof a circular arc as described above, Rl
(Ar)
= rl
(AR )
.
Now, with this theorem, one can prove the fundamental inequality of Corollary
2.3.10.
Proof:Situate the angle θ such that one side is on the positive x axis and extend the
other side till it intersects the unit circle at the point
(cosθ,sinθ)
. Then denoting the
resulting arc on the circle by A, it follows that for all P ∈P
(A)
the inequality
(1 − cosθ)
+ sinθ ≥
|P |
≥ sinθ. It follows that
(1− cosθ)
+ sinθ is an upper bound for
all the
|P|
where P ∈P
(A )
and so
(1 − cosθ)
+ sinθ is at least as large as the sup or
least upper bound of the
|P |
. This proves the top half of the inequality. The bottom half
follows because l
(A )
≥ L where L is the length of the line segment joining
(cosθ,sinθ)
and
(1,0)
due to the definition of l
(A)
. However, L ≥ sinθ because L is the
length of the hypotenuse of a right triangle having sinθ as one of the sides.
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