2
sin
sin
1
+
+
+
+
.
56*. Yig’indilarni toping:
(
)
x
n
x
x
1
2
sin
...
3
sin
sin
2
2
2
−
+
+
+
.
57. Isbotlang:
a)
(
)
x
in
s
nx
in
s
x
n
s
co
n
nx
s
co
x
s
co
x
s
co
2
1
2
...
2
2
2
2
+
+
=
+
+
+
;
b)
(
)
inx
s
nx
in
s
x
n
s
co
n
nx
in
s
x
in
s
x
in
s
2
1
2
...
2
2
2
2
+
−
=
+
+
+
.
58*. Yig’indilarni toping:
a)
nx
n
x
x
x
cos
...
3
cos
3
2
cos
2
cos
+
+
+
+
;
b)
nx
n
x
x
x
sin
...
3
sin
3
2
sin
2
sin
+
+
+
+
.
5-§. Birning ildizlari
Har qanday noldan farqli kompleks son kabi 1 sonning ham n–darajali
ildizi n ta qiymatga ega.
0
sin
0
cos
1
i
+
=
bo’lganligi uchun 1 ning n
−
darajali
ildizlari uchun
n
in
s
i
n
s
co
πκ
πκ
ε
κ
2
2
+
=
,
1
....,
,
1
,
0
−
=
n
k
formula o’rinli.
1 ning n
−
darajali ildizi boshlang’ich ildiz deyiladi, agar u 1 ning n dan
kichik darajali ildizi bo’lmasa. Boshqacha aytganda,
ε
son 1 ning n
−
darajali
boshlang’ich ildizi bo’ladi, agar
1
=
n
ε
bo’lib, istagan
n
m
<
uchun
1
m
≠
ε
bo’lsa.
n
in
s
i
n
s
co
π
π
ε
2
2
1
+
=
sonning 1 ning n
−
darajali boshdang’ich ildizi
bo’lishi ravshan, lekin n > 2 bo’lgandan undan boshqa boshlang’ich ildizlar
ham mavjud.
1-m i s o l.
n
in
s
i
n
s
co
πκ
πκ
ε
2
2
+
=
(
2
,
1
≥
≥
n
k
-butun sonlar) son 1 ning
d
n
darajali boshlang’ich ildizi bo’lishini ko’rsating, bu yerda d son k va n
larning eng katta umumiy bo’luvchisidan iborat. Bu yerdan kelib chiqadiki,
ε
son 1 ning n
−
darajali boshlang’ich ildizi bo’lishi uchun k va n larning o’zaro
tub bo’lishi zarur va yetarlidir.
25
Yechish.
d
n
n
d
1
1
,
=
=
κ
κ
bo’lsin, bu yera n
1
va
κ
1
o’zaro tub sonlar. U
holda
1
1
1
1
2
sin
2
cos
n
i
n
πκ
πκ
ε
+
=
.
ε
ni
0
,
1
>
<
m
n
m
darajaga ko’tarib,
1
1
1
1
2
2
n
m
in
s
i
n
m
s
co
m
κ
π
κ
π
ε
+
=
ni hosil qilamiz.
Agar
1
=
m
ε
bo’lsa, u holda
s
n
m
π
κ
π
2
2
1
1
=
, bunda
Z
∈
s
yoki
s
n
m
=
1
1
κ
,
ya’ni
m
1
κ
son
1
n ga bo’linadi. Lekin
1
κ
va
1
n o’zaro tub. Shuning uchun m
son
1
n ga bo’linadi, bu esa 0 < m <
1
n shartga zid.
Shunday qilib, 0 < m <
1
n
bo’lganda
1
≠
m
ε
. m =
1
n bo’lganda esa
1
2
sin
2
cos
1
1
1
=
+
=
πκ
πκ
ε
i
n
. Bu yerdan
ε
- son 1 ning
d
n
n
=
1
darajali
boshlang’ich ildizi ekanligi kelib chiqadi. ■
Bu misoldan ko’rinadiki, 1 ning n-darajali boshlang’ich ildizlari soni n
dan kichik va n bilan o’zaro tub bo’lgan sonlar soniga, ya’ni Eyler
funksiyasining
( )
n
ϕ
qiymatiga tengdir.
( )
(
)
( )
∏
=
−
=
n
k
n
x
x
X
ϕ
κ
ε
1
ko’phad, bu yerda
κ
ε
(
( )
n
ϕ
κ
...,
,
1
,
0
=
) - 1 ning
boshlang’iya ildizi, doiraviy ko’phad deyiladi.
2-m i s o l. Birning 6-darajali ildizlarini toping.
Yechilishi: 1 ning n-darajali ildizlari formulasidan:
5.
0,1,2,3,4,
=
+
=
к
,
6
к
2
isin
6
к
2
cos
е
к
π
π
ni hosil qilamiz. Natijada, izlanayotgan ildizlar quyidagilardan iborat bo’ladi:
1
0
0
=
+
=
in
s
i
s0
co
ε
,
i
i
2
3
2
1
3
sin
3
cos
1
+
=
+
=
π
π
ε
,
i
i
2
3
2
1
3
2
sin
3
2
cos
2
+
−
=
+
=
π
π
ε
,
1
sin
cos
3
−
=
+
=
π
π
ε
i
,
i
i
2
3
2
1
3
4
sin
3
4
cos
4
−
−
=
+
=
π
π
ε
,
i
i
2
3
2
1
3
5
sin
3
5
cos
5
−
=
+
=
π
π
ε
. ■
Agar 1 ning n-darajali ildizi 1 ning
δ
darajali boshlang’ich ildizi bo’lsa,
δ
son bu ildiz tegishli bo’lgan ko’rsatkich deyiladi.
3-m i s o l. Birning 6-darajali boshlang’ich ildizlarini yozing.
Yechish. 1-misolga ko’ra birning 6-darajali boshlang’ich ildizlari
5
,
1
ε
ε
lardan, ya’ni
i
2
3
2
1
±
lardan iborat. ■
4-m i s o l.
0
1
5
=
−
x
tenglamani algebraik yo’l bilan yechib, 1 ning 5-
darajali ildizlarini toping.
Yechish. Tenglamaning chap tomonini ko’paytuvchilarga ajratamiz:
26
(
)
(
)
0
1
1
2
3
4
=
+
+
+
+
−
x
x
x
x
x
.
Boshlang’ich ildizlar quyidagi tenglamaning ildizlari bo’ladi:
0
1
2
3
4
=
+
+
+
+
x
x
x
x
.
Bu tenglama
0
1
1
2
2
=
+
+
+
+
−
−
x
x
x
x
tenglamaga teng kuchli.
1
−
+
=
x
x
z
belgilash kiritamiz.
2
2
2
2
z
x
x
=
+
+
−
bo’lishini e’tiborga olsak,
0
1
2
=
−
+
z
z
tenglama hosil bo’ladi, bundan
2
5
2
1
1
+
−
=
z
,
2
5
2
1
2
−
−
=
z
.
1 ning ildizlari
0
1
1
2
=
+
−
x
z
x
va
0
1
2
2
=
+
−
x
z
x
tenglamalardan
topiladi. Bulardan
2
4
2
1
1
z
i
z
x
−
±
=
va
2
4
2
2
2
z
i
z
x
−
±
=
. z
1
va z
2
larning
qiymatlarini qo’yib,
4
5
2
10
4
1
5
1
+
+
−
=
i
x
,
4
5
2
10
4
1
5
2
+
−
−
=
i
x
,
4
5
2
10
4
1
5
3
−
+
−
−
=
i
x
,
4
5
2
10
4
1
5
4
−
−
−
−
=
i
x
larni hosil qilamiz.
Bularni
0
0
72
72
5
2
5
2
⋅
+
⋅
=
+
=
κ
κ
πκ
πκ
κ
isin
s
co
in
s
i
s
co
x
formula bilan
taqqoslab
4
1
5
72
cos
0
−
=
ni hosil qilamiz.
Bundan r radiusli doiraga ichki chizilgan muntazam o’nburchakning
tomoni a
10
uchun formula keltirib chiqariladi:
r
r
s
rco
in
s
r
in
s
r
a
⋅
−
=
=
−
=
=
⋅
=
2
1
5
5
2
cos
2
10
2
2
10
2
2
10
2
2
10
π
π
π
π
π
.■
5-m i s o l. Birning 7 ko’rsatkichga tegishli bo’lgan 28-darajali ildizlarini
yozing.
Yechish. Ma’lumki, 1 ning 28-darajali ildizlari
(
)
27
,....,
2
,
1
,
0
,
28
2
28
2
=
+
=
κ
πκ
πκ
ε
in
s
i
s
co
k
,
lardan iborat. Bulardan 7 ko’rsatkichga tegishli bo’lganlari
,
,
,
,
,
,
24
20
16
12
8
4
ε
ε
ε
ε
ε
ε
lardir yoki ularni quyidagicha yozish mumkin:
6
,
5
,
4
,
3
,
2
,
1
,
0
бунда
,
7
2
7
2
=
+
κ
πκ
πκ
in
s
i
s
co
. ■
6-m i s o l.
1
−
n
x
ni haqiqiy koeffisiyentli ko’paytuvchilarga ajrating.
Yechish.
m
n
2
=
bo’lsin, u holda
0
1
=
−
n
x
tenglama ikkita 1, -1 haqiqiy
ildizlarga va
2
2
−
m
ta kompleks ildizlarga ega.
Bunda
m
in
s
i
m
s
co
2
2
2
2
πκ
πκ
ε
κ
+
=
ildiz
27
(
)
(
)
m
m
in
s
i
m
m
s
co
m
2
2
2
2
2
2
2
π
κ
π
κ
ε
κ
−
+
−
=
−
ildizga qo’shma.
Shunday qilib,
(
)
(
)
(
)
(
)
(
)
(
)
1
1
2
2
1
1
2
2
...
1
1
−
−
−
−
−
−
−
−
−
=
−
m
m
m
x
x
x
x
x
x
x
x
ε
ε
ε
ε
ε
ε
;
(
)
(
)
[
]
(
)
[
]
1
...
1
1
1
1
1
2
1
1
2
2
2
+
+
−
+
+
−
−
=
−
−
−
x
x
x
x
x
x
m
m
m
ε
ε
ε
ε
;
(
)
+
−
−
=
−
Π
−
=
1
2
1
1
2
1
1
2
2
m
s
xco
x
x
x
m
m
κπ
κ
.
ni hosil qilamiz.
Agar n = 2m+1 bo’lsa, shunga o’xshash yo’l bilan
(
)
+
+
−
−
=
−
Π
=
+
1
1
2
2
2
1
1
2
1
1
2
m
s
xco
x
x
x
m
m
κπ
κ
ni hosil qilamiz. ■
7-m i s o l. Tenglikni isbotlang:
(
)
1
2
2
1
...
2
2
2
−
=
−
⋅
m
m
m
m
in
s
m
in
s
m
in
s
π
π
π
.
Yechish. 6-misolning natijasiga ko’ra:
+
−
=
−
−
Π
−
=
1
cos
2
1
1
2
1
1
2
2
m
x
x
x
x
m
m
κπ
κ
ni hosil qilamiz.
x = 1 ni qo’yib,
−
=
Π
−
=
−
m
s
co
m
m
m
κπ
κ
1
2
1
1
1
yoki
(
)
m
in
s
m
m
m
2
2
2
1
1
1
2
κπ
κ
Π
−
=
−
=
va
nihoyat
m
in
s
m
m
m
2
2
1
1
1
κπ
κ
Π
−
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