143
26- § da (3-masala)
sin(
)
cos , cos(
)
sin
2
2
p
p
a
a
a
a
-
=
-
=
formulalar isbotlangan edi, ular ham
keltirish formulalari
deb ataladi.
Bu
formulalardan foydalanib, masalan, sin
cos ,
p
p
3
6
=
cos
sin
p
p
3
6
=
ni
hosil qilamiz.
x
ning istalgan qiymati uchun sin(
x
+
2
p
)
=
sin
x
, cos(
x
+
2
p
)
=
cos
x
tengliklar to‘g‘riligi ma’lum.
Bu tengliklardan ko‘rinadiki, argument 2
p
ga o‘zgarganda sinus
va kosinusning qiymatlari davriy takrorlanadi. Bunday funksiyalar
davri
2
p
bo‘lgan davriy funksiyalar
deyiladi.
Agar shunday
T
¹
0 son mavjud bo‘lsaki,
y
=
f
(
x
) funksiya-
ning aniqlanish sohasidagi istalgan
x
uchun
f
(
x
-
T
)
=
f
(
x
)
=
f
(
x
+
T
)
tenglik bajarilsa,
f
(
x
)
davriy funksiya deb
ataladi.
T
son
f
(
x
) funksiyaning
davri
deyiladi.
Bu ta’rifdan ko‘rinadiki, agar
x
son
f
(
x
) funksiyaning aniqlanish
sohasiga tegishli bo‘lsa, u holda
x
+
T
,
x
-
T
sonlar va, umuman,
x
+
Tn
,
n
Î
Z
sonlar ham shu davriy funksiyaning aniqlanish sohasiga tegishli
va
f
(
x
+
Tn
)
=
f
(
x
),
n
Î
Z
bo‘ladi.
2
p
soni
y
=
cos
x
funksiyaning
eng kichik musbat davri
eka-
nini ko‘rsatamiz.
T
>
0 kosinusning davri bo‘lsin, ya’ni
istalgan
x
uchun
cos(
x
+
T
)
=
cos
x
tenglik bajariladi.
x
=
0 deb, cos
T
=
1 ni hosil qilamiz.
Bundan esa
T
=
2
p
k
,
k
Î
Z
. T
>
0 bo‘lganidan
T
quyidagi 2
p
, 4
p
, 6
p
, ...
qiymatlarni qabul qila oladi va shuning uchun
T
ning qiymati 2
p
dan
kichik bo‘lishi mumkin emas.
y
=
sin
x
funksiyaning eng kichik musbat davri ham
2
p
ga
teng
ekanini isbotlash mumkin.
!
!
145
359.
Ifodani soddalashtiring:
1)
( )
sin
sin(
)
cos(
) sin(
)
p a
p a
p a
p a
2
2
- +
-
- +
-
;
2)
( )
( )
cos(
) cos
sin(
) sin
p a
p a
p a
p a
- +
-
- -
-
2
2
;
3)
(
)
( )
sin
tg(
)
tg(
)
cos
a p
a p
p a
p a
-
+
×
-
-
2
;
4)
( )
sin (
) sin
sin(
)
tg(
)
2
2
2
p a
p a
p a
p a
- +
-
-
×
-
.
360.
Uchburchakning ikkita ichki burchagi yig‘indisining sinusi
uchinchi burchagining sinusiga tengligini isbotlang.
361.
Ayniyatni isbotlang:
1)
( )
sin
cos
p
a
a
2
+
=
;
2)
( )
cos
sin
p
a
a
2
+
= -
;
3)
(
)
cos
sin
3
2
p a
a
-
= -
;
4)
(
)
sin
cos
3
2
p a
a
-
= -
.
362.
Tenglamani yeching:
1)
( )
cos
p
2
1
-
=
x
;
2)
(
)
sin
p -
=
x
1 ;
3)
(
)
cos
x
-
=
p
0 ;
4)
( )
sin
x
-
=
p
2
1.
29- §.
SINUSLAR YIG‘INDISI VA AYIRMASI.
KOSINUSLAR YIG‘INDISI VA AYIRMASI
1 - m a s a l a .
Ifodani soddalashtiring:
( )
( )
(
)
sin
sin
sin
a
a
p
p
p
+
+
-
12
12
12
.
Qo‘shish formulasi va ikkilangan
burchak sinusi formulasidan
foydalanib, quyidagiga ega bo‘lamiz:
( )
( )
(
)
sin
sin
sin
a
a
p
p
p
+
+
-
=
12
12
12
(
)
=
+
+
-
=
sin cos
cos sin
sin cos
cos sin
sin
a
a
a
a
p
p
p
p
p
12
12
12
12
12
=
×
=
=
2
12
12
6
1
2
sin cos
sin
sin sin
sin
a
a
a
p
p
p
.
10 – Algebra, 9- sinf uchun
146
Agar
sinuslar yig‘indisi formulasi
sin
sin
sin
cos
a
b
a b
a b
+
=
+
-
2
2
2
(1)
dan foydalanilsa, bu masalani soddaroq yechish mumkin. Shu formula
yordamida quyidagini hosil qilamiz:
( )
( )
(
)
sin
sin
sin
a
a
p
p
p
+
+
-
=
12
12
12
=
×
=
2
12
12
1
2
sin cos
sin
sin
a
a
p
p
.
Endi (1) formulaning o‘rinli ekanini isbotlaymiz.
a b
a b
+
-
=
=
2
2
x
y
,
belgilash kiritamiz. U holda
x
+
y
=
a
,
x
-
y
= b
va
shuning uchun sin
a +
sin
b
=
sin(
x
+
y
)
+
sin(
x
-
y
)
=
sin
x
cos
y
+
cos
x
sin
y
+
+
sin
x
cos
y
-
cos
x
sin
y
=
2sin
x
cos
y
=
+
-
2
2
2
sin
cos
a b
a b
.
(1) formula bilan bir qatorda quyidagi
sinuslar ayirmasi formulasi,
kosinuslar yig‘indisi
va
ayirmasi formulalaridan
ham foydalaniladi:
sin
sin
sin
cos
a
b
a b
a b
-
=
-
+
2
2
2
,
(2)
cos
cos
cos
cos
a
b
a b
a b
+
=
+
-
2
2
2
,
(3)
cos
cos
sin
sin
a
b
a b
a b
-
= -
+
-
2
2
2
.
(4)
(3) va (4) formulalar ham (1) formulaning isbotlanishiga o‘xshash
isbotlanadi; (2) formula
b
ni
-b
ga almashtirish bilan (1) formuladan
hosil qilinadi (
buni mustaqil isbotlang
).
2 - m a s a l a .
sin75
° +
cos75
°
ni hisoblang.
sin75
° +
cos75
° =
sin75
° +
sin15
° =
=
=
=
+
-
2
2
45
30
75
15
2
75
15
2
sin
cos
sin
cos
o
o
o
o
o
o
2
6
2
3
2
2
2
=
×
.
147
3- m a s a l a .
2
3
sin
a +
ni ko‘paytmaga almashtiring.
(
)
(
)
2
3
2
2
3
2
3
sin
sin
sin
sin
a
a
a
p
+
=
+
=
+
=
(
)
(
)
=
+
-
4
2
6
2
6
sin
cos
a
p
a
p
.
4- m a s a l a .
sin
a +
cos
a
ifodaning
eng kichik qiymati
-
2 ga,
eng katta qiymati esa 2 ga teng ekanini isbotlang.
Berilgan ifodani ko‘paytmaga almashtiramiz:
(
)
(
)
sin
cos
sin
sin
sin cos
a
a
a
a
a
p
p
p
+
=
+
-
=
-
=
2
4
4
2
(
)
4
cos
2
p
-
a
.
Kosinusning eng kichik qiymati
-
1 ga, eng katta qiymati esa 1 ga
teng bo‘lgani uchun berilgan ifodaning eng kichik qiymati 2
1
× - =
( )
= -
2 ga, eng katta qiymati esa 2 1
2
× =
ga teng.
M a s h q l a r
363.
Ifodani soddalashtiring:
1) sin(
(
p
p
a
a
3
3
+
) sin
)
+
-
;
2) cos(
cos(
p
p
b
b
4
4
-
-
+
)
) ;
3) sin (
(
2
2
4
4
p
p
a
a
+
) sin
)
-
-
; 4) cos (
cos (
2
2
4
4
a
a
p
p
-
-
+
)
) .
364.
Hisoblang:
1) cos
cos
105
75
o
o
+
;
2) sin
sin
105
75
o
o
-
;
3) cos
11
12
5
12
p
p
+
cos
;
4) cos
11
12
5
12
p
p
-
cos
;
5) sin
cos
7
12
12
p
p
-
;
6)
o
o
165
sin
105
sin
+
.
365.
Ko‘paytmaga almashtiring:
1) 1
+
2sin
a
;
2) 1
-
2sin
a
;
3) 1
+
2cos
a
;
4) 1
+
sin
a
.
366.
Ayniyatni isbotlang:
1)
sin
sin
cos
cos
tg
a
a
a
a
a
+
+
=
3
3
2 ;
2)
sin
sin
cos
cos
ctg
2
4
2
4
a
a
a
a
a
+
-
=
.
