Download Solutions to Exercises, Section 6.1

Instructor’s Solutions Manual, Section 6.1
Exercise 1
Solutions to Exercises, Section 6.1
1.
Find the area of a triangle that has sides of length 3 and 4, with an
angle of 37◦ between those sides.
◦
3·4·sin 37
, which equals
solution The area of this triangle equals
2
◦
6 sin 37 . A calculator shows that this is approximately 3.61 (make sure
that your calculator is computing in degrees, or first convert to radians,
when doing this calculation).
Instructor’s Solutions Manual, Section 6.1
2.
Exercise 2
Find the area of a triangle that has sides of length 4 and 5, with an
angle of 41◦ between those sides.
◦
41
solution This triangle has area 4·5·sin
, which equals 10 sin 41◦ . A
2
calculator shows that this is approximately 6.56 (make sure that your
calculator is computing in degrees, or first convert to radians, when
doing this calculation).
Instructor’s Solutions Manual, Section 6.1
3.
Exercise 3
Find the area of a triangle that has sides of length 2 and 7, with an
angle of 3 radians between those sides.
3
solution The area of this triangle equals 2·7·sin
, which equals 7 sin 3.
2
A calculator shows that this is approximately 0.988 (make sure that
your calculator is computing in radians, or first convert to degrees,
when doing this calculation).
Instructor’s Solutions Manual, Section 6.1
4.
Exercise 4
Find the area of a triangle that has sides of length 5 and 6, with an
angle of 2 radians between those sides.
2
solution The area of this triangle equals 5·6·sin
, which equals
2
15 sin 2. A calculator shows that this is approximately 13.6395 (make
sure that your calculator is computing in radians, or first convert to
degrees, when doing this calculation).
Instructor’s Solutions Manual, Section 6.1
Exercise 5
For Exercises 5–12 use the following figure (which is not drawn to scale):
a
Θ
b
5. Find the value of b if a = 3, θ = 30◦ , and the area of the triangle equals
5.
solution Because the area of the triangle equals 5, we have
5=
ab sin θ
2
=
3b sin 30◦
2
=
Solving the equation above for b, we get b =
3b
4 .
20
3 .
Instructor’s Solutions Manual, Section 6.1
Exercise 6
6. Find the value of a if b = 5, θ = 45◦ , and the area of the triangle equals
8.
solution Because the area of the triangle equals 8, we have
=
√
5a 2
4 .
Solving the equation above for a, we get a =
√
16 2
5 .
8=
ab sin θ
2
=
a5 sin 45◦
2
Instructor’s Solutions Manual, Section 6.1
7. Find the value of a if b = 7, θ =
10.
Exercise 7
π
4,
and the area of the triangle equals
solution Because the area of the triangle equals 10, we have
10 =
ab sin θ
2
=
7a sin
2
π
4
=
Solving the equation above for a, we get a =
7a
√ .
2 2
√
20 2
7 .
Instructor’s Solutions Manual, Section 6.1
8. Find the value of b if a = 9, θ =
π
3,
Exercise 8
and the area of the triangle equals 4.
solution Because the area of the triangle equals 4, we have
=
√
9b 3
4 .
Solving the equation above for b, we get b =
√
16 3
27 .
4=
ab sin θ
2
=
9b sin
2
π
3
Instructor’s Solutions Manual, Section 6.1
9.
Exercise 9
Find the value of θ (in radians) if a = 7, b = 6, the area of the triangle
π
equals 15, and θ < 2 .
solution Because the area of the triangle equals 15, we have
15 =
ab sin θ
2
=
7·6·sin θ
2
= 21 sin θ.
5
Solving the equation above for sin θ, we get sin θ = 7 . Thus
5
θ = sin−1 7 ≈ 0.7956.
Instructor’s Solutions Manual, Section 6.1
10.
Exercise 10
Find the value of θ (in radians) if a = 5, b = 4, the area of the triangle
π
equals 3, and θ < 2 .
solution Because the area of the triangle equals 3, we have
3=
ab sin θ
2
=
5·4·sin θ
2
= 10 sin θ.
Solving the equation above for sin θ, we get sin θ =
3
θ = sin−1 10 ≈ 0.3047.
3
10 .
Thus
Instructor’s Solutions Manual, Section 6.1
11.
Exercise 11
Find the value of θ (in degrees) if a = 6, b = 3, the area of the
triangle equals 5, and θ > 90◦ .
solution Because the area of the triangle equals 5, we have
5=
ab sin θ
2
=
6·3·sin θ
2
= 9 sin θ.
5
Solving the equation above for sin θ, we get sin θ = 9 . Thus θ equals
5
π − sin−1 9 radians. Converting this to degrees, we have
◦
θ = 180◦ − (sin−1 95 ) 180
≈ 146.25◦ .
π
Instructor’s Solutions Manual, Section 6.1
12.
Exercise 12
Find the value of θ (in degrees) if a = 8, b = 5, and the area of the
triangle equals 12, and θ > 90◦ .
solution Because the area of the triangle equals 12, we have
12 =
ab sin θ
2
=
8·5·sin θ
2
= 20 sin θ.
3
Solving the equation above for sin θ, we get sin θ = 5 . Thus θ equals
3
π − sin−1 5 radians. Converting this to degrees, we have
◦
θ = 180◦ − (sin−1 53 ) 180
≈ 143.13◦ .
π
Instructor’s Solutions Manual, Section 6.1
13.
Exercise 13
Find the area of a parallelogram that has pairs of sides of lengths 6
and 9, with an angle of 81◦ between two of those sides.
solution The area of this parallelogram equals 6 · 9 · sin 81◦ , which
equals 54 sin 81◦ . A calculator shows that this is approximately 53.34.
Instructor’s Solutions Manual, Section 6.1
14.
Exercise 14
Find the area of a parallelogram that has pairs of sides of lengths 5
and 11, with an angle of 28◦ between two of those sides.
solution The area of this parallelogram equals 5 · 11 · sin 28◦ , which
equals 55 sin 28◦ . A calculator shows that this is approximately 25.82.
Instructor’s Solutions Manual, Section 6.1
Exercise 15
15. Find the area of a parallelogram that has pairs of sides of lengths 4 and
π
10, with an angle of 6 radians between two of those sides.
solution The area of this parallelogram equals 4 · 10 · sin π6 , which
equals 20.
Instructor’s Solutions Manual, Section 6.1
Exercise 16
16. Find the area of a parallelogram that has pairs of sides of lengths 3 and
π
12, with an angle of 3 radians between two of those sides.
solution The area of this parallelogram equals 3 · 12 · sin π3 , which
√
equals 18 3.
Instructor’s Solutions Manual, Section 6.1
Exercise 17
For Exercises 17–24, use the following figure (which is not drawn to scale
except that u is indeed meant to be an acute angle and ν is indeed
meant to be an obtuse angle):
b
a
a
Ν
u
b
17. Find the value of b if a = 4, ν = 135◦ , and the area of the parallelogram
equals 7.
solution Because the area of the parallelogram equals 7, we have
√
7 = ab sin ν = 4b sin 135◦ = 2 2b.
Solving the equation above for b, we get b =
7
√
2 2
=
√
7 2
4 .
Instructor’s Solutions Manual, Section 6.1
Exercise 18
18. Find the value of a if b = 6, ν = 120◦ , and the area of the parallelogram
equals 11.
solution Because the area of the parallelogram equals 11, we have
√
11 = ab sin ν = a6 sin 120◦ = 3 3a.
Solving the equation above for a, we get a =
11
√
3 3
=
√
11 3
9 .
Instructor’s Solutions Manual, Section 6.1
19. Find the value of a if b = 10, u =
equals 7.
Exercise 19
π
3,
and the area of the parallelogram
solution Because the area of the parallelogram equals 7, we have
√
7 = ab sin u = 10a sin π3 = 5a 3.
Solving the equation above for a, we get a =
7
√
5 3
=
√
7 3
15 .
Instructor’s Solutions Manual, Section 6.1
20. Find the value of b if a = 5, u =
equals 9.
π
4,
Exercise 20
and the area of the parallelogram
solution Because the area of the parallelogram equals 9, we have
9 = ab sin u = 5b sin π4 =
Solving the equation above for b, we get b =
√
5b 2
2 .
18
√
5 2
=
√
9 2
5 .
Instructor’s Solutions Manual, Section 6.1
21.
Exercise 21
Find the value of u (in radians) if a = 3, b = 4, and the area of the
parallelogram equals 10.
solution Because the area of the parallelogram equals 10, we have
10 = ab sin u = 3 · 4 · sin u = 12 sin u.
5
Solving the equation above for sin u, we get sin u = 6 . Thus
u = sin−1
5
6
≈ 0.9851.
Instructor’s Solutions Manual, Section 6.1
22.
Exercise 22
Find the value of u (in radians) if a = 4, b = 6, and the area of the
parallelogram equals 19.
solution Because the area of the parallelogram equals 19, we have
19 = ab sin u = 4 · 6 · sin u = 24 sin u.
Solving the equation above for sin u, we get sin u =
u = sin−1
19
24
≈ 0.9135.
19
24 .
Thus
Instructor’s Solutions Manual, Section 6.1
23.
Exercise 23
Find the value of ν (in degrees) if a = 6, b = 7, and the area of the
parallelogram equals 31.
solution Because the area of the parallelogram equals 31, we have
31 = ab sin ν = 6 · 7 · sin ν = 42 sin ν.
31
Solving the equation above for sin ν, we get sin ν = 42 . Because ν is an
31
obtuse angle, we thus have ν = π − sin−1 42 radians. Converting this to
degrees, we have ν = 180◦ − (sin−1
31 180 ◦
42 ) π
≈ 132.43◦ .
Instructor’s Solutions Manual, Section 6.1
24.
Exercise 24
Find the value of ν (in degrees) if a = 8, b = 5, and the area of the
parallelogram equals 12.
solution Because the area of the parallelogram equals 12, we have
12 = ab sin ν = 8 · 5 · sin ν = 40 sin ν.
3
Solving the equation above for sin ν, we get sin ν = 10 . Because ν is an
3
obtuse angle, we thus have ν = π − sin−1 10 radians. Converting this to
degrees, we have ν = 180◦ − (sin−1
3 180 ◦
10 ) π
≈ 162.54◦ .
Instructor’s Solutions Manual, Section 6.1
Exercise 25
25. What is the largest possible area for a triangle that has one side of
length 4 and one side of length 7?
solution In a triangle that has one side of length 4 and one side of
length 7, let θ denote the angle between those two sides. Thus the area
of the triangle will equal
14 sin θ.
We need to choose θ to make this area as large as possible. The largest
π
possible value of sin θ is 1, which occurs when θ = 2 (or θ = 90◦ if we
π
are working in degrees). Thus we choose θ = 2 , which gives us a right
triangle with sides of length 4 and 7 around the right angle.
This right triangle has area 14, which is the
largest area of any triangle with sides of
length 4 and 7.
4
7
Instructor’s Solutions Manual, Section 6.1
Exercise 26
26. What is the largest possible area for a parallelogram that has pairs of
sides with lengths 5 and 9?
solution In a parallelogram that has pairs of sides with lengths 5 and
9, let θ denote an angle between two adjacent sides. Thus the area of
the parallelogram will equal
45 sin θ.
We need to choose θ to make this area as large as possible. The largest
π
possible value of sin θ is 1, which occurs when θ = 2 (or θ = 90◦ if we
π
are working in degrees). Thus we choose θ = 2 , which gives us a
rectangle with sides of length 5 and 9.
This rectangle has area 45, which is the
largest area of any parallelogram with sides of
length 5 and 9.
5
9
Instructor’s Solutions Manual, Section 6.1
Exercise 27
27. Sketch the regular hexagon whose vertices are six equally spaced points
on the unit circle, with one of the vertices at the point (1, 0).
solution
1
Instructor’s Solutions Manual, Section 6.1
Exercise 28
28. Sketch the regular dodecagon whose vertices are twelve equally spaced
points on the unit circle, with one of the vertices at the point (1, 0).
[A dodecagon is a twelve-sided polygon.]
solution
1
Instructor’s Solutions Manual, Section 6.1
Exercise 29
29. Find the coordinates of all six vertices of the regular hexagon whose
vertices are six equally spaced points on the unit circle, with (1, 0) as
one of the vertices. List the vertices in counterclockwise order starting
at (1, 0).
solution The coordinates of the six vertices, listed in
2π m
2π m
counterclockwise order starting at (1, 0), are (cos 6 , sin 6 ), with
m going from 0 to 5. Evaluating the trigonometric √functions,√we get the
1
3
1
3
following list of coordinates of vertices: (1, 0), ( 2 , 2 ), (− 2 , 2 ),
1
(−1, 0), (− 2 , −
√
3
2 ),
( 12 , −
√
3
2 ).
Instructor’s Solutions Manual, Section 6.1
Exercise 30
30. Find the coordinates of all twelve vertices of the dodecagondodecagon
whose vertices are twelve equally spaced points on the unit circle, with
(1, 0) as one of the vertices. List the vertices in counterclockwise order
starting at (1, 0).
solution The coordinates of the twelve vertices, listed in
2π m
2π m
counterclockwise order starting at (1, 0), are (cos 12 , sin 12 ), with
m going from 0 to 11. Evaluating the trigonometric √
functions, √we get
3 1
1
3
the following list of coordinates of vertices: (1, 0), ( 2 , 2 ), ( 2 , 2 ),
√
3
3 1
),
(−
2 , 2 ),
√2
( 23 , − 12 ).
1
(0, 1), (− 2 ,
√
1
3
( 2 , − 2 ),
√
√
(−1, 0), ( −2 3 , − 12 ), (− 12 , −
√
3
2 ),
(0, −1),
Instructor’s Solutions Manual, Section 6.1
Exercise 31
31. Find the area of a regular hexagon whose vertices are six equally spaced
points on the unit circle.
solution Decompose the hexagon into triangles by drawing line
segments from the center of the circle (the origin) to the vertices. Each
triangle has two sides that are radii of the unit circle; thus those two
sides of the triangle each have length 1. The angle between those two
2π
radii is 6 radians (because one rotation around the entire circle is an
angle of 2π radians, and each of the six triangles has an angle that
2π
π
takes up one-sixth of the total). Now 6 radians equals 3 radians (or
60◦ ). Thus each of the six triangles has area
1
2
which equals
6·
√
3
4 ,
√
3
4 .
· 1 · 1 · sin π3 ,
Thus the sum of the areas of the six triangles equals
which equals
√
3 3
2 .
In other words, the hexagon has area
√
3 3
2 .
Instructor’s Solutions Manual, Section 6.1
Exercise 32
32. Find the area of a regular dodecagon whose vertices are twelve equally
spaced points on the unit circle.
solution Decompose the dodecagon into triangles by drawing line
segments from the center of the circle (the origin) to the vertices. Each
triangle has two sides that are radii of the unit circle; thus those two
sides of the triangle each have length 1. The angle between those two
2π
radii is 12 radians (because one rotation around the entire circle is an
angle of 2π radians, and each of the twelve triangles has an angle that
2π
π
takes up one-twelfth of the total). Now 12 radians equals 6 radians (or
30◦ ). Thus each of the twelve triangles has area
1
2
1
· 1 · 1 · sin π6 ,
which equals 4 . Thus the sum of the areas of the twelve triangles
1
equals 12 · 4 , which equals 3. In other words, the dodecagon has area 3.
Instructor’s Solutions Manual, Section 6.1
Exercise 33
33. Find the perimeter of a regular hexagon whose vertices are six equally
spaced points on the unit circle.
solution If we assume that one of the vertices of the hexagon is the
point (1, 0), then
the next vertex in the counterclockwise direction is
√
1
3
the point ( 2 , 2 ). Thus the length of each side of the hexagon equals
1
the distance between (1, 0) and ( 2 ,
#
1−
√
3
2 ),
1 2
2
+
which equals
√3 2
2
,
which equals 1. Thus the perimeter of the hexagon equals 6 · 1, which
equals 6.
Instructor’s Solutions Manual, Section 6.1
Exercise 34
34. Find the perimeter of a regular dodecagondodecagon whose vertices
are twelve equally spaced points on the unit circle.
solution If we assume that one of the vertices of the dodecagon is
the point (1, 0),
√ then the next vertex in the counterclockwise direction
3 1
is the point ( 2 , 2 ). Thus the length of each side of the dodecagon
√
equals the distance between (1, 0) and (
#
1−
which
equals
√
12 2 − 3.
√
3 2
2
+
3 1
2 , 2 ),
1 2
2
which equals
,
√
2 − 3. Thus the perimeter of the dodecagon equals
Instructor’s Solutions Manual, Section 6.1
Exercise 35
35. Find the area of a regular hexagon with sides of length s.
solution There is a constant c such that a regular hexagon with sides
2
of length s has
√ area cs . From Exercises 31 and 33, we know that the
3 3
area equals 2 if s = 1. Thus
√
3 3
2
= c · 12 = c.
Thus a regular hexagon with sides of length s has area
√
3 3 2
2 s .
Instructor’s Solutions Manual, Section 6.1
Exercise 36
36. Find the area of a regular dodecagon with sides of length s.
solution There is a constant c such that a regular dodecagon with
2
. From Exercises 32 and 34, we know that
sides of length s has area
cs √
the area equals 3 if s = 2 − 3. Thus
3=c
√ 2
√
2 − 3 = c(2 − 3).
Solving this equation for c, we have
√
√
3
2+ 3
3
√ =
√ ·
√ = 6 + 3 3.
c=
2− 3
2− 3 2+ 3
√
Thus a regular dodecagon with sides of length s has area (6 + 3 3)s 2 .
Instructor’s Solutions Manual, Section 6.1
37.
Exercise 37
Find the area of a regular 13-sided polygon whose vertices are 13
equally spaced points on a circle of radius 4.
solution Decompose the 13-sided polygon into triangles by drawing
line segments from the center of the circle to the vertices. Each triangle
has two sides that are radii of the circle with radius 4; thus those two
sides of the triangle each have length 4. The angle between those two
2π
radii is 13 radians (because one rotation around the entire circle is an
angle of 2π radians, and each of the 13 triangles has an angle that takes
up one-thirteenth of the total). Thus each of the 13 triangles has area
1
2
· 4 · 4 · sin
2π
13 ,
2π
which equals 8 sin 13 . The area of the 13-sided polygon is the sum of
2π
the areas of the 13 triangles, which equals 13 · 8 sin 13 , which is
approximately 48.3.
Instructor’s Solutions Manual, Section 6.1
38.
Exercise 38
The face of a Canadian one-dollar coin is a regular 11-sided polygon
(see the picture just before the start of these exercises). The distance
from the center of this polygon to one of the vertices is 1.325
centimeters. Find the area of the face of this coin.
solution Think of the face of the coin as being inscribed in a circle
with radius 1.325 centimeters. Decompose the 11-sided polygon into
triangles by drawing line segments from the center of the circle to the
vertices. Each triangle has two sides that are radii of the circle with
radius 1.325 centimeters; thus those two sides of the triangle each have
2π
length 1.325 centimeters. The angle between those two radii is 11
radians (because one rotation around the entire circle is an angle of 2π
radians, and each of the 11 triangles has an angle that takes up
one-eleventh of the total). Thus each of the 11 triangles has area
1
2
· 1.325 · 1.325 · sin
2π
11 ,
square centimeters. The area of the 11-sided polygon is the sum of the
1
2π
areas of the 11 triangles, which equals 11 · 2 · 1.325 · 1.325 · sin 11
square centimeters, which is approximately 5.22 square centimeters.
Instructor’s Solutions Manual, Section 6.1
Problem 39
Solutions to Problems, Section 6.1
39. What is the area of a triangle whose sides all have length r ?
solution A triangle all of whose sides have length r is an equilateral
π
triangle all of whose angles are 3 radians (or 60◦ ). Thus the area of
such a triangle is
1 2
2r
sin
π
3,
√
which equals
3r 2
4 .
Instructor’s Solutions Manual, Section 6.1
Problem 40
40. Explain why there does not exist a triangle with area 15 having one side
of length 4 and one side of length 7.
solution Consider a triangle with one side of length 4, one side of
length 7, and an angle θ between these two sides. The area of this
1
triangle equals 2 · 4 · 7 sin θ, which equals 14 sin θ. Because sin θ ≤ 1,
the area of this triangle is less than or equal to 14. Thus this triangle
cannot have area 15.
Instructor’s Solutions Manual, Section 6.1
Problem 41
41. Show that if a triangle has area R, sides of length A, B, and C, and
angles a, b, and c, then
1
R 3 = 8 A2 B 2 C 2 (sin a)(sin b)(sin c).
[Hint: Write three formulas for the area R, and then multiply these
formulas together.]
solution Consider a triangle with sides of length A, B, and C. Let a
be the angle between the sides of length B and C, let b be the angle
between the sides of length A and C, and let c be the angle between the
sides of length A and B. Let R be the area of this triangle. Then we have
the formulas
1
R = 2 BC sin a
R = 12 AC sin b
R = 12 AB sin c.
Multiplying these three formulas together gives
1
R 3 = 8 A2 B 2 C 2 (sin a)(sin b)(sin c).
Instructor’s Solutions Manual, Section 6.1
Problem 42
42. Find numbers b and c such that an isosceles triangle with sides of
length b, b, and c has perimeter and area that are both integers.
solution Consider an isosceles triangle with sides of length b, b, and
c. The perimeter of this triangle is 2b + c.
Let h denote the height of this triangle as shown in the figure below.
The area of this triangle is ch
2 . As can be seen in the figure below, we
c
have a right triangle with sides of length h, 2 , and b. To make
everything in sight an integer, we will make this be the right triangle
with sides of length 3, 4, and 5. In other words, we choose b = 5 and
c
c = 8 (which makes 2 = 4 and, by the Pythagorean Theorem, makes
h = 4).
b
b
h
c2
With b = 5 and c = 8, this triangle has perimeter 18 and area 20.
Of course there are also other correct solutions.
Instructor’s Solutions Manual, Section 6.1
Problem 43
43. Explain why the solution to Exercise 32 is somewhat close to π .
solution The area inside a circle of radius 1 is π , which is
approximately 3.14. The regular dodecagon in Exercise 32 fills up most
of the area inside the unit circle, so its area should be just somewhat
less that the area inside the circle. Indeed, the dodecagon in Exercise 32
has area 3, so it misses about 0.14 of the area inside the circle.
Instructor’s Solutions Manual, Section 6.1
44.
Problem 44
Use a calculator to evaluate numerically the exact solution you
obtained to Exercise 34. Then explain why this number is somewhat
close to 2π .
solution In Exercise 34 we found that a dodecagon whose vertices
√
are equally spaced points on the unit circle has perimeter 12 2 − 3. A
calculator shows that this number is approximately 6.21.
The perimeter of this dodecagon should be just somewhat less than the
circumference of the unit circle, as can be seen in the figure that is the
solution to Exercise 28. The unit circle has circumference 2π , which is
approximately 6.28. Thus the perimeter of the dodecagon
(approximately 6.21) is somewhat close to the circumference of the
circle (approximately 6.28) as we indeed expect from the figure.
Instructor’s Solutions Manual, Section 6.1
Problem 45
45. Explain why a regular polygon with n sides whose vertices are n
n
2π
equally spaced points on the unit circle has area 2 sin n .
solution Consider a regular polygon with n sides whose vertices are
n equally spaced points on the unit circle. Draw a radius from the
origin to each vertex, partitioning the polygon into n isosceles
triangles. Each of these n isosceles triangles has two sides of length 1
2π
(the radii), with angle n between these two sides (because the entire
circle, which has angle 2π , has been partitioned into n equal angles).
1
2π
Thus the area of each of the n isosceles triangles is 2 · 1 · 1 · sin n ,
which equals
1
2
sin
2π
n .
Because the regular polygon with n sides is composed of n of these
n
2π
triangles, the area of the polygon equals 2 sin n .
Instructor’s Solutions Manual, Section 6.1
Problem 46
46. Explain why the result stated in the previous problem implies that
sin
2π
n
≈
2π
n
for large positive integers n.
solution Suppose n is a large positive integer. Then a regular
polygon with n sides whose vertices are n equally spaced points on the
unit circle fills up almost all the area inside the unit circle. Thus the
n
2π
area of this polygon, which from the previous problem equals 2 sin n ,
should be approximately the same as the area inside the unit circle,
which equals π . In other words, we have
n
2
Multiplying both sides by
2
n
sin
2π
n
≈ π.
gives the approximation
sin 2π
n ≈
2π
n .
Instructor’s Solutions Manual, Section 6.1
47.
Problem 47
Choose three large values of n, and use a calculator to verify that
2π
2π
sin n ≈ n for each of those three large values of n.
2π
2π
solution The table below gives the values of sin n and n to six
significant digits for n = 1000, n = 10000, and n = 100000:
n
1000
10000
100000
sin 2π
n
2π
n
0.00628314
0.00628319
0.000628318
0.000628319
0.0000628319
0.0000628319
2π
As can be seen from the table above, we have sin 2π
n ≈ n for these
three large values of n, with the approximation more accurate for
larger values of n.
Instructor’s Solutions Manual, Section 6.1
Problem 48
48. Show that each edge of a regular polygon with n sides whose vertices
are n equally spaced points on the unit circle has length
2 − 2 cos 2π
n .
solution Consider a regular polygon with n sides whose vertices are
n equally spaced points on the unit circle, with one of the vertices at
the point (1, 0). The next vertex in the counterclockwise direction has
2π
2π coordinates cos n , sin n , because the radius ending at that vertex
has angle
2π
n
with the positive horizontal axis.
Thus the length of this edge of the polygon (and hence of each edge of
the polygon) is the distance between the points (1, 0) and
2π
2π cos n , sin n . We can compute that distance using the usual formula
for the distance between two points. Thus the length of each edge of
this polygon is given by the following formula:
#
1 − cos
2π 2
n
+ sin2
2π
n
2π
2π
2π
= 1 − 2 cos n + cos2 n + sin2 n
= 2 − 2 cos 2π
n
Instructor’s Solutions Manual, Section 6.1
Problem 49
49. Explain why a regular polygon with n sides, each with length s, has area
2π
n
s2.
cos 2π
)
n
n sin
4(1 −
solution A regular polygon with n sides, with the vertices equally
n
2π
spaced on the unit circle, has area 2 sin n (from Problem 45) and has
sides of length 2 − 2 cos 2π
n (from Problem 48).
Thus to get a regular polygon with sides having length s, we
horizontally and vertically stretch this polygon with vertices on the unit
circle by a factor of
s
.
2 − 2 cos 2π
n
By the Area Stretch Theorem (Section 4.2), this changes the area by a
factor of
2
s
.
2 − 2 cos 2π
n
Thus a regular polygon with n sides, each with length s, has area
n
2π
2 (sin n )
which equals
s
2 − 2 cos 2π
n
2
,
Instructor’s Solutions Manual, Section 6.1
n sin 2π
n
4(1 − cos 2π
n )
Problem 49
s2.
Instructor’s Solutions Manual, Section 6.1
Problem 50
50. Verify that for n = 4, the formula given by the previous problem
reduces to the usual formula for the area of a square.
solution If n = 4, then
n sin 2π
n
4(1 − cos 2π
n )
s2 =
4 sin π2
4(1 − cos
π s
2)
2
= s2,
which is the usual formula for the area of a square with sides of
length s.
Instructor’s Solutions Manual, Section 6.1
Problem 51
51. Explain why a regular polygon with n sides whose vertices are n
equally spaced points on the unit circle has perimeter
n 2 − 2 cos 2π
n .
solution By Problem 48, we know that each side of a regular polygon
with
n sides whose vertices lie on the unit circle has length
2 − 2 cos 2π
n . The perimeter of the polygon is n times the length of
each side. Thus the perimeter of this polygon is
n 2 − 2 cos 2π
n .
Instructor’s Solutions Manual, Section 6.1
Problem 52
52. Explain why the result stated in the previous problem implies that
n 2 − 2 cos 2π
n ≈ 2π
for large positive integers n.
solution If n is a large positive integer, then the perimeter of regular
polygon with n sides whose vertices lie on the unit circle is
approximately equal to the circumference of the unit circle, which
equals 2π . In other words, if n is a large positive integer then
2π
n 2 − 2 cos n ≈ 2π .
Instructor’s Solutions Manual, Section 6.1
53.
Problem 53
Choose three large values of n, and use a calculator to verify that
2π
n 2 − 2 cos n ≈ 2π for each of those three large values of n.
solution The table below gives the values of n 2 − 2 cos 2π
n and 2π
to six significant digits for n = 100, n = 1000, and n = 10000:
n 2 − 2 cos 2π
n
2π
100
6.28215
6.28319
1000
6.28317
6.28319
10000
6.28319
6.28319
n
As can be seen from the table above, we have n 2 − 2 cos 2π
n ≈ 2π for
these three large values of n, with the approximation more accurate for
larger values of n.
Instructor’s Solutions Manual, Section 6.1
Problem 54
54. Show that
cos
2π
n
≈1−
2π 2
n2
if n is a large positive integer.
solution Suppose n is a large positive integer. Then we know from
Problem 52 that
2π
n 2 − 2 cos n ≈ 2π .
Square both sides of the approximation above and then divide by n2 to
get
2π
4π 2
2 − 2 cos n ≈ n2 .
Now divide both sides by 2 and then solve for cos 2π
n , getting
cos
2π
n
≈1−
2π 2
n2 .