Trigonometry+ 09: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 09


Problem A

Prove:
lim_x→0[sin(x)/x] = 1

Proof A
In the lecture Trigonometry 07 (Problem C) of this Math+ & Problems course we have proven geometrically that for an acute angle θ, measured in radians, the following inequalities are true:
sin(θ) ≤ θ ≤ tan(θ)
The left inequality can be transformed into
sin(x)/x ≤ 1
Using the definition of function tan(x)=sin(x)/cos(x), the right inequality can be transformed into
cos(x) ≤ sin(x)/x
Combining together, these inequalities give
cos(x) ≤ sin/x ≤ 1
As x→0, lower and upper boundaries of sin(x)/x tend to 1.
Therefore, lim_x→0[sin(x)/x] = 1

A different proof can be obtained based on using the L'Hopital's Rule (see UNIZOR.COM: Math 4 Teens - Calculus - Derivatives - Main Theorems - L'Hopital's Rule) as follows:
lim_x→0[sin(x)/x] =
= lim_x→0[sin'(x)/x'] =
= lim_x→0[cos(x)/1] = 1


The following theorems assume that the overall dimensions of regular polygons are not growing to infinity but restricted to some values, while the number of sides of polygons increases to infinity.


Problem B

Prove that the perimeter of a regular polygon tends to a circumference of an inscribed into it circle, if the number of its sides tends to infinity.
More precisely, prove that the difference between a perimeter of a regular N-sided polygon and a circumference of a circle inscribed into it is an infinitesimal variable as N→∞.

Proof B
Let
p - perimeter of a regular polygon,
N - number of its sides,
r - radius of an inscribed circle.
Then,
p = 2N·r·tan(½·2π/N) =
= 2N·r·sin(π/N)/cos(π/N) =
(substitute x=π/N,
so if N→∞, x→0)
= 2π·r·[sin(x)/x]/cos(x)
As x→0, cos(x)→1.
As proven in Problem A above, lim_x→0[sin(x)/x] = 1
Therefore, the limit of a perimeter p equals to 2π·r, which is a circumference of an inscribed circle.


Problem C

Prove that the perimeter of a regular polygon tends to a circumference of a circumscribed around it circle, if the number of its sides tends to infinity.
More precisely, prove that the difference between a perimeter of a regular N-sided polygon and a circumference of a circle circumscribed around it is an infinitesimal variable as N→∞.

Proof C
Let
p - perimeter of a regular polygon,
N - number of its sides,
R - radius of a circumscribed circle.
Then,
p = 2N·R·sin(½·2π/N) =
= 2N·R·sin(π/N) =
(substitute x=π/N,
so if N→∞, x→0)
= 2π·R·[sin(x)/x]
As x→0, sin(x)/x→1.
Therefore, the limit of a perimeter p equals to 2π·R - a circumference of a circumscribed circle.


Problem D

Prove that the area of a regular polygon tends to an area of an inscribed into it circle, if the number of its sides tends to infinity.
More precisely, prove that the difference between an area of a regular N-sided polygon and an area of a circle inscribed into it is an infinitesimal variable as N→∞.

Proof D
Let
A - area of a regular polygon,
p - perimeter of a regular polygon,
N - number of its sides,
r - radius of an inscribed circle.
Dividing a regular polygon into N equilateral triangles with common vertex in a center of an inscribed circle leads to an obvious formula for its area
A = ½·p·r
Since the limit of a perimeter p equals to 2π·r,
lim_N→∞A = ½·2π·r·r = π·r²


Problem E

Prove that the area of a regular polygon tends to an area of a circumscribed around it circle, if the number of its sides tends to infinity.
More precisely, prove that the difference between an area of a regular N-sided polygon and an area of a circle circumscribed around it is an infinitesimal variable as N→∞.

Proof E
Let
A - area of a regular polygon,
p - perimeter of a regular polygon,
N - number of its sides,
R - radius of a circumscribed circle.
r - radius of an inscribed circle.
Notice that
r² + (p/2N)² = R²
and, therefore,
lim_N→∞(R−r) = 0.
That is, radiuses of inscribed and circumscribed circles have the same limit.
Dividing a regular polygon into N equilateral triangles with common vertex in a center of a circumscribed circle leads to an obvious formula for its area
A = ½·p·√R²−(p/2N)²
In this formula, as proved above, the limit of a perimeter p equals to 2π·r, which is the same as 2π·R.
An expression p/2N has limit zero as N→∞.
Therefore,
lim_N→∞A = ½·2π·R·R = π·R²