19
44*.
nx
in
s
n
va
nx
s
co
n
larni (n – butun musbat son) x ga karali
burchaklarning sinusi va kosinuslarining birinchi darajali ko’phadi ko’rinishida
ifodalash mumkinligini isbotlang.
45. x ga karrali burchaklar trigonometrik funksiyalarining birinchi darajali
ko’phadlari ko’rinishida tasvirlang:
a)
x
in
s
3
; b)
x
in
s
4
; c)
x
s
co
5
;
d)
x
s
co
6
; e )
x
s
co
x
in
s
5
3
; f)
x
sin
x
cos
7
7
+
.
4-§. Yig’indi va ko’paytmalarni kompleks
sonlar yordamida hisoblash
1-m i s o l. Ayniyatni isbotlang:
1
3
1
4
2
2
..
...
1
−
=
+
+
=
+
+
+
n
n
n
n
n
C
C
C
C
.
Yechish. Nyuton binomini qo’lab quyidagi tengliklarni hosil qilamiz:
( )
0
)
1
1
(
)
1
...(
1
,
2
1
1
...
1
3
2
1
1
3
2
1
=
−
=
−
+
−
+
−
=
+
=
+
+
+
+
+
+
−
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
C
C
C
C
C
C
C
C
C
0
)
1
1
(
)
1
...(
1
3
2
1
=
−
=
−
+
−
+
−
n
n
n
n
n
n
n
C
C
C
C
.
Bu tengliklarni hadlab qo’shib, undan keyin ayirib, kerakli ayniyatni hosil
qilamiz. ■
2-m i s o l. Ayniyatni isbotlang:
+
=
+
+
+
3
рn
2cos
2
3
1
....
C
C
C
n
6
n
3
n
0
n
.
Yechish. Quyidagi tenglikni qaraymiz:
(
)
n
n
n
n
n
n
n
n
n
n
n
x
C
x
C
x
C
x
C
x
C
C
x
+
+
+
+
+
+
=
+
−
−
1
1
3
3
2
2
1
0
...
1
.
Bu tenglikga ketma-ket x = 1,
2
,
ε
ε
larni qo’yamiz, bu yerda
0
1
2
=
+
+
ε
ε
. Natijada quyidagi tengliklar hosil bo’ladi:
(
)
(
)
...
1
...,
1
...,
2
6
3
4
3
2
1
0
2
3
3
2
2
1
0
3
2
1
0
+
+
+
+
=
+
+
+
+
+
=
+
+
+
+
+
=
ε
ε
ε
ε
ε
ε
ε
ε
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
C
C
C
C
C
C
C
C
C
C
C
C
Lekin k son 3 ga bo’linmaganda
0
1
2
=
+
+
κ
κ
ε
ε
, k son 3 ga bo’linganda
esa
3
1
2
=
+
+
κ
κ
ε
ε
bo’ladi. Shuning uchun yuqoridagi tengliklarni hadlab
qo’shib,
(
)
(
) {
}
...
3
1
1
2
6
3
0
2
+
+
+
=
+
+
+
+
n
n
n
n
n
n
C
C
C
ε
ε
tenglikni hosil qilamiz.
3
2
sin
3
2
cos
π
π
ε
i
+
=
deb olish mumkin bo’lganligi uchun
20
.
3
3
3
2
3
2
1
,
3
3
3
4
3
4
1
2
2
π
π
π
π
ε
ε
π
π
π
π
ε
ε
in
s
i
s
co
in
s
i
s
co
in
s
i
s
co
in
s
i
s
co
−
=
−
−
=
−
=
+
+
=
−
−
=
−
=
+
Shuning uchun
(
)
(
)
3
cos
2
2
1
1
2
2
π
ε
ε
n
n
n
n
n
+
=
+
+
+
+
. Bu yerdan
+
=
+
+
+
3
cos
2
2
3
1
...
6
3
0
π
n
C
C
C
n
n
n
n
. ■
3-m i s o l. Tenglikni isbotlang:
(
)
2
2
2
1
2
2
1
2
2
2
...
2
...
2
1
C
C
C
C
n
n
n
n
+
+
+
=
+
+
+
−
+
.
Yechish.
(
)
2
2
2
1
1
2
2
κ
κ
κ
κ
κ
−
=
⋅
−
=
C
bo’lganligi uchun
κ
κ
κ
−
=
2
2
2C
.
Shuning uchun
∑
∑
∑
=
=
=
−
=
n
n
n
k
k
C
2
2
2
2
2
2
κ
κ
κ
κ
.
Bundan yuqoridagi ayniyat kelib chiqadi. ■
4-m i s o l. Yig’indini hisoblang:
...
1
6
4
2
+
−
+
−
=
n
n
n
C
C
C
δ
Yechish.
( )
....
1
1
3
3
2
2
1
+
+
+
+
=
+
i
C
i
C
i
C
i
n
n
n
n
ifodani qaraymiz. Bundan
( )
(
) (
)
....
...
1
1
5
3
1
4
2
−
+
−
+
−
+
−
=
+
n
n
n
n
n
n
C
C
C
i
C
C
i
.
Lekin
( )
+
=
+
4
sin
4
cos
2
1
π
π
i
i
. Shuning uchun
4
cos
2
...
1
2
6
4
2
π
δ
n
C
C
C
n
n
n
n
=
+
−
+
−
=
.
Bu yerdan
n = 4
m bo’lganda
( )
m
m
2
2
1
−
=
δ
, n = 4m+1 bo’lganda
( )
m
m
2
2
1
−
=
δ
, n = 4m+3 bo’lganda
( )
1
2
1
2
1
+
+
−
=
m
m
δ
, n = 4m+2 bo’lganda
0
=
δ
bo’lishi kelib chiqadi. ■
5-m i s o l. Ayniyatni isbotlang:
.
1
1
2
1
1
...
3
1
2
1
1
1
2
1
+
−
=
+
+
+
+
+
+
n
C
n
C
C
n
n
n
n
n
Yechish. Quyidagi ko’phadni qaraymiz:
(
)
1
1
1
2
2
1
1
1
1
...
1
1
+
+
+
+
+
+
+
+
+
=
+
n
n
n
n
n
n
x
C
x
C
x
C
x
.
Bundan
(
)
1
3
2
2
1
0
1
1
...
2
2
1
1
1
+
+
+
+
+
+
+
=
+
−
+
n
n
n
n
n
n
n
x
n
C
x
C
x
C
x
C
n
x
(*).
Bu tenglikka x = 1 ni qo’yib, izlanayotgan ayniyatni hosil qilamiz. ■
6-m i s o l. Ayniyatni isbotlang:
1
1
2
1
...
+
+
+
+
+
+
=
+
+
+
+
n
n
n
n
n
n
n
n
n
n
C
C
C
C
C
κ
κ
.
Yechish. Tenglikning chap tomonidagi ifoda
21
(
) (
)
(
)
(
)
κ
+
+
+
+
+
+
+
+
+
+
+
=
n
n
n
n
x
x
x
x
S
1
...
1
1
1
2
1
ko’phaddagi
x
n
oldidagi koeffisiyentdan iborat. Bu ko’phadni quyidagicha
almashtiramiz:
(
)
(
) (
)
(
)
[
]
(
) (
)
(
)
(
)
[
]
.
1
1
1
1
1
1
1
...
1
1
1
1
1
1
2
n
n
n
n
x
x
x
x
x
x
x
x
x
x
S
+
−
+
=
=
−
+
+
=
+
+
+
+
+
+
+
+
=
+
+
+
κ
κ
κ
kvadrat qavs ichidagi ko’phadda
1
+
n
x
oldidagi koeffisiyent
1
1
+
+
+
n
n
C
κ
ga teng
bo’ladi. ■
7-m i s o l.
p
n
m
m
p
n
p
m
n
p
m
n
C
C
C
C
C
C
C
+
−
=
+
+
+
0
1
1
0
...
tenglik o’rinli bo’lishini
ko’rsating.
Yechish. Quyidagi ko’phadlarni qaraymiz:
(
) ∑
=
=
+
n
s
s
s
n
n
x
C
x
0
1
,
(
) ∑
=
=
+
m
t
t
t
m
m
x
C
x
0
1
.
U holda:
(
) (
)
(
)
∑
∑
∑
+
+
=
Do'stlaringiz bilan baham: