Zahiriddin muhammad bobur nomidagi andijon davlat universiteti fizika-matematika fakulteti
§5. Ikki burchak yig’indisi va ayirmasining trigonometrik funksiyalari
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§5. Ikki burchak yig’indisi va ayirmasining trigonometrik funksiyalari.
Ikkilangan va uchlangan burchakning trigonometrik funksiyalari Ba’zi
hollarda berilgan burchakning trigonometrik funksiyalari qiymatlarini bevosita hisoblab bo’lmaydi. Lekin bu burchakni ikkita burchak yig’indisi yoki ayirmasi sifatida qaralib quyidagi formulalardan foydalanilsa ularni bevosita hisoblash mumkin bo’ladi. Ular:
( ) 2 3 cos(
α π α π − • − tg ) ( ) 2 3 cos( α π α π − • −
α • α
α α •
α ) 2 ( cos
) ( cos 2 2
x + + + π π ) 2 ( cos ) ( cos 2 2 x x + + + π π • • • • • • • • • • • • • • •
• ) 2 3 ( ) ( ) cos( ) 2 sin( α π α π α π α π − • + + • −
ctg ) 2 3 ( ) ( ) cos( ) 2 sin( α π α π α π α π − • + + • −
ctg = − = • − • = α α α α α α 2 2 cos
) cos
( cos
ctg ctg ctg = − α α α 2 2 2 sin cos
cos α β α β α β α sin cos cos
sin ) sin( • + • = + β α β α β α sin cos cos
sin ) sin( • − • = −
- 22 -
Misollar: 1. cos45 0 cos15 0 +sin45
0 sin15
0 ni hisoblang. Yechish: cos45 0 cos15 0 +sin45
0 sin15
0 =cos(40
0 -15
0 )=cos30
0 =
Javob:
2. Agar va bo’lsa,
ning qiymatini toping.
Yechish: =cos
cos -sin
sin +2sin
sin =cos
cos +sin
sin =cos(
- )=cos(-45 0 -15
0 )=cos(-60 0 )=cos60
0 =
Javob: 3.
ni soddalashtiring Yechish:
= . 4. ni hisoblang. Yechish: 2(sin30
0 cos15
0 + +cos30 0 sin15
0 )= 2 sin(30 0 +15
0 )=2sin45
0 =2 = . 5. sin105 0 +sin75
0 =?
β α β α β α sin sin
cos cos
) cos(
• − • = + β α β α β α sin sin cos
cos ) cos( • − • = + β α β α β α
tg tg tg tg • − + = + 1 ) ( β α β α β α tg tg tg tg tg • + − = − 1 ) ( β α β α β α tg tg tg tg ctg + • − = + 1 ) ( β α β α β α tg tg tg tg ctg − • + = − 1 ) ( • • • • 3 3 3 3 0 45 − = α 0 15 = β β α β α sin sin
2 ) cos( • + + β α β α sin
sin 2 ) cos( • + + α •
β α •
β α •
β α •
β α •
β α β 2 1 2 1 β α β α β α β α cos sin
2 ) sin( sin sin
2 ) cos( • − + • + + β α β α β α β α cos sin 2 ) sin( sin
sin 2 ) cos( • − + • + + = • − • + • • + • − • = β α β α β α β α β α β α cos
sin 2 sin cos cos
sin sin
sin 2 sin sin cos
cos = • + • • + • β α β α β α β α sin
cos cos
sin sin
sin cos
cos ) ( ) sin(
) cos(
) sin(
) cos(
α β α β α β α β β α − = − − = − −
0 0
sin 3 15 cos + 0 0 15 sin 3 15 cos + = + = ) 15 sin 2 3 15 cos
2 1 ( 2 0 0 • • • • 2 2 2 - 23 -
Yechish: sin105 0 +sin75
0 =sin(60
0 +45
0 )+sin(45
0 +30
0 )=sin60
0 cos45
0 + +cos60 0 sin45
0 + sin45
0 cos30
0 +cos45
0 sin30
0 =
=
= . 6. ni soddalashtiring. Yechish:
0 . 7. tg105 0 ni hisoblang. Yechish: tg105 0 =tg(60 0 +45
0 )=
= . 8. Agar va bo’lsa, ning qiymatlarini toping.
Yechish: , , . 9.
bo’lsa, tg ning qiymatini toping. • •
• = + + + 4 2 4 6 4 2 4 6 = + = + = • + • 2 2 6 2 2 2 6 4 2 2 4 6 2 = + = + 2 1 3 2 ) 1 3 ( 2 = + + = + 2 1 3 2 3 2 ) 1 3 ( 2 = + = + = 2 ) 3 2 ( 2 2 3 2 4 3 2 +
0 0 0 0 0 0 0 0 168 sin 108
sin 78 sin 18 sin
208 sin
108 cos
28 cos
18 cos
• + • • + • = • + • • + • 0 0 0 0 0 0 0 0 168 sin
108 sin
78 sin
18 sin
208 sin
108 cos
28 cos
18 cos
= − • + + − • + • + + • = ) 12 180 sin(
) 18 90 sin( ) 12 90 sin(
18 sin
) 28 180 sin( ) 18 90 cos(
28 cos
18 cos
0 0 0 0 0 0 0 0 0 0 0 0 0 = + • + • = ) 12 18 sin(
18 sin
28 sin
18 cos
28 cos
0 0 0 0 0 0 = = − 2 1 10 cos 30 sin ) 18 28 cos( 0 0 0 0 = − + = • − + 3 1 1 3 45 60 1 45 60 0 0 0 0 tg tg tg tg = + − + + = − + ) 3 1 )( 3 1 ( ) 3 1 )( 3 1 ( 3 1 3 1 = + − = − + + 2 3 2 4 3 1 3 3 2 1 3 2 ) 3 2 ( − − = + − 1 cos sin = • β α 2 1 cos
sin = • α β β α − 2 1 2 1 1 sin
cos cos
sin ) sin( = − = • − • = − β α β α β α 2 1 ) sin( = − β α π π β α
k + • − = − 6 ) 1 ( 2 ) 4 ( = − α π tg α
- 24 -
Yechish: , 1-tg
=2+2tg ;
3tg =-1; tg
= . Ba’zi hollarda burchak trigonometrik funksiyalarining qiymatlarini bilgan holda 2 va 3 burchakning trigonometrik funksiyalari qiymatlarini hisoblashga to’g’ri keladi. U quyidagi formulalar yordamida amalga oshiriladi. sin2
=2sin cos
; cos2 =cos
2 -sin
2 ; tg2
= ; ctg
=
sin3 =3sin -4sin
3 ; cos3 =4cos 3
; tg3 =
ctg3 =
Bu formulalarni barchasini ikki burchak yig’indisining trigonometrik funksiyalarini berilgan burchaklar trigonometrik funksiyalari orqali ifodalash formulalaridan foydalanib keltirib chiqariladi. cos2
=cos 2 -sin 2 formulaning o’ng tomonidagi cos 2 ni sin
2 orqali
yoki sin 2 ni cos 2 orqali ifodalab quyidagi formulalarni hosil qilish mumkin. cos2 =1-2sin
2 yoki cos2 =2cos 2
Bu formulalardan esa yoki
formulalarni hosil qilamiz. Bu formulalarni trigonometrik Funksiyalarni darajalarini pasaytirish formulalari deb ham ataladi. Yuqorida ko’rib o’tilgan formulalardan 2 ni
bilan almashtirib yana bir qancha formulalarni hosil qilish mumkin. , ,
,
; 1 1 4 1 4 ) 4 ( α α α π α π α π tg tg tg tg tg tg tg + − = • + − = − 2 1 1 = + − α α
tg α α α α 3 1 − α α α α α • α α α α α α α 2 1 2 tg tg − α α α
tg 2 1 2 − α α α α α α α α α α 2 3 3 1 3
tg tg − − α α α α 3 3 3 3 1 tg tg tg − − α α α α α α α α α α 2 2 cos 1 sin 2 α α − = 2 2 cos
1 cos
2 α α + = α α 2 cos 2 sin
2 sin
α α α • = 2 sin 2 cos cos 2 2 α α α − = 2 1 2 2 2 α α α tg tg tg − = 2 2 2 1 2 α α α
tg ctg − = - 25 -
Bu formulalar trigonometrik ifodalarni soddalashtirishda, trigonometrik tenglamalar va tengsizliklarni yechishda ko’p qo’llaniladi. Misollar: 1. ni hisoblang Yechish:
2. ni hisoblang. Yechish:
3. ni soddalashtiring. Yechish:
sin +cos -sin
=cos
4. Yechish:
5. cos92
0 cos2
0 +0,5sin4
0 +1 ni hisoblang. Yechish: cos92 0 cos2 0 +0,5sin4
0 +1=cos(90 0 +2
) cos2 0 +0,5sin4 0 +1=
=-sin2 0 cos2 0 + 2sin2 0 sos2
0 +1=-sin2
0 cos2+sin2 0 cos2
0 +1=1.
6. ni soddalashtiring Yechish: 3ctg
2 . ) 8 3 cos 8 3 (sin 2 14 4 4 π π − ) 8 3 cos
8 3 (sin 2 14 4 4 π π − ) 8 3 cos
8 3 (sin 2 14 2 2 π π + = = − ) 8 3 cos
8 3 (sin 2 2 π π = − − • • = )) 8 3 sin
8 3 (cos ( 1 2 14 2 2 π π = • • − 8 3 2 cos 2 14 π = − 4 3 cos 2 14 π = + • − = ) 4 2 cos( 2 14 π π 14 2 2 2 14 ) 4 sin ( 2 14 = • = − • − π 8 9 cos 8 7 sin 8 2 2 π π • 8 9 cos 8 7 sin 8 2 2 π π • = + • − = ) 8 ( cos ) 8 ( sin
8 2 2 π π π π = • • 8 cos 8 sin
8 2 2 π π = • • = 2 ) 8 cos 8 sin 2 ( 2 π π 1 4 2 2 ) 2 2 ( 2 4 sin 2 2 2 = • = • = π α α α α sin cos sin
2 sin
1 − + + α α α α sin cos sin
2 sin
1 − + + = − + + • + = α α α α α α α sin cos
sin cos
cos sin
2 sin
2 2 = − + + = α α α α α sin cos
sin ) cos (sin 2 α α α α ? 10 cos 50 sin
40 sin
4 0 0 0 = • = • • = • 0 0 0 0 0 0 10 cos 40 cos 40 sin
2 2 10 cos 50 sin 40 sin
4 2 10 cos 10 cos 2 10 cos 80 sin
2 0 0 0 0 = = • • • • • 2 1 •
• • • α α α 2 2 sin cos 2 cos 1 + + α α α 2 2 sin cos 2 cos 1 + + = = + = α α α α α 2 2 2 2 2 sin cos 3 sin cos cos
2 α
- 26 -
7. ni soddalashtiring. Yechish: cos -2
8. ni hisoblang. Yechish:
. 9. ni qiymatini toping. Yechish:
=
10. bo’lsa, tg2 ni qiymatini hisoblang. Yechish: .
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