Thiết kế bài giảng đại số và giải tích 11 nâng cao (tập 1) phần 1

TRAN VINH

TAP

MOT

NHA XUAT BAN HA NOI



TRAN VINH

^

^

>.

THIET KE BAI GIANG
DAI SO VA GIAI TICH

/\

/\

NANG CAO - TAP MOT

NHA XUAT BAN HA NO!


Thiet kebai giang

OAI SO VA GIAI TICH 11 - TAP MOT, NANG CAO
TRAN VINH

N H A XUAT BAN H A N O I

Chiu trdch nhiem xudt bdn:
NGUYfiN K H A C O A N H
Bien tap:
PHAM QUOC TUAN
Vebia:
NGUYfiN T U X N
Trinh bdy:
THAI SON - SON LAM
Svta bdn in:
PHAM QUOC TUAN

In 1.000 cuon khd 17 x 24 cm tai Cong ty TNHH Bao bi va In Hai Nam.
Quyet dinh xuat ban so: 127 - 2008/CXB/100'= TK - 05/HN.
In xong va nop liiu chieu quy III nam 2O08.


J^ai noi adu
Trong nhi5ng nam gin day, thuc hidn ddi mdi chuefng trinh Sach giao
khoa (SGK) cua Bo Giao duo va Dao tao, b6 SGK mdri ra dcfi, trong do co b6
sach bien scan theo chuong trinh phan ban cua bac Trung hoc pho thong. Bo
sach gom ba ban: Ban co ban. Ban nang cao khoa hoc tu nhien va Ban nang
cao khoa hoc xa hoi.
Viec ra bo sach SGK m6i ddng nghia vdi vi6c phai ddi m6i phuong phap
day va hoc. Nham dap img nhiing yeu cSu do, tiep ndi bo sach: ThiSt k6' bai
giang m6n toan ldp 10, chiing toi tiep tuc bien scan bo sach: Thiet ke bai

giang mon Toan ldp 11.
Bg sach gom, 8 cuon:
Thiet kebdi gidng Hinh hoc 11:2 tap
Thiei kebdi gidng Dgi sdvd Gidi tich 11:2 tap
Thiit kebdi gidng Hinh hoc 11 ndng cao: 2 tap
Thiit kebdi gidng Dgi sdvd Gidi tich 11 ndng cao: 2 tap
Day la b6 sach co nhieu hudng thiet ke, co nhieu dang, nhieu loai cau
hoi, bai tap nham hudng hoc sinh (HS) den nhung don vi kien thiic nha't
dinh. He thdng cac cau hoi trac nghiem khach quan d cudi bai nham giup HS
on tap va nang cao ki nang phan doan, quy nap, tijr dd xac dinh duoc noi
dung kidn thiic chu yeu va co ban ciia bai hoc.
Bo sach duoc cac tac gia cd nhieu kinh nghiem trong giang day, trong
nghien cuu khoa hoc (dac biet cd nhieu tac gia da nghien ciiu nhung phSn
mem de hd tro trong giang day, nha't la cac mdn hoc khoa hoc tu nhifen, toan
hoc...). Bien soan bg sach ra ddi hy vong giiip ban doc cd mdt each nhin
mdi, phucmg phap mdi. Cac each thiet ke trong bo sach nay vtta cd tmh dinh
hudng, vita cu the, nham tao ra cac hudng md de giao vien (GV) ap dung ddi
vdi nhiing ddi tugng HS khac nhau.
Tuy da nghien ciiu va bien scan cin than, soiig khdng the tranh nhiing sai
sdt, tac gia kfnh mong dugc su gdp y ciia ban dgc.
TAC GIA



CHlTdNG I
HAM SO Ll/dNG GIAC
•

VA PHlTdNG TRINH LlTdNG GIAC
^

•

PHan 1

^sHwf^G TAX: n& CUA cm/di^G
I. NOI DUNG
Ngi dung chinh cua chuefng 1:
Ham sd lugng giac : Tinh tuSn hoan, su bien thien cua cac ham sd
y = sinx, y = cosx, y = tanx va y = cotx.
Phucfng trinh lugng giac co ban : Cdng thiic nghiem va didu kien cd
nghiem cua cac phucfng trinh sinx = m, cosx = m, tanx = m va cotx = m.
Dac biet la chii y de'n cac phuofng trinh sinx = sin a, cosx = cosa,
tanx = tana va cotx = cota.
Mgt sd phucfng trinh lugng giac thudng gap: Phuong trinh dua ve bac nh&,
bac hai ddi vdi cac ham sd lugng giac; Phucfng trinh bac nhit ddi vdi sinx
va cosx, phucmg trinh thuan nhat bac hai ddi vdi sinx va cosx va mgt sd
dang phuong trinh khac.
II. MUCTIEU
1. Kien thurc
Nam dugc toan bd kien thiic co ban trong chucmg da neu tren, cu the :
Hieu khai niem, chieu bie'n thien, tinh tudn hoan cua cac ham sd lugng giac.
Ap dung chieu bien thien va tinh tuan hoan ciia cac ham sd lugng giac de
giai dugc cac phucmg trinh lugng giac.
Nam dugc cac cdng thiic nghiem dl giai cac phuong trinh lugng giac ccf ban.


Hilu each tim nghiem cua cac phuong trinh lugng giac ccf ban va phuomg
phap giai mdt sd dang phucmg trinh lugng giac don gian.
Nam dugc mdt sd phuomg phap giai mdt sd dang phucmg trinh lugng

giac khac.
Hieu khai niem cac ham sd lugng giac y =sinx, y = cosx, y = tanx,
y = cot X va tinh chat tuSn hoan cua chung.
Nam dugc su bie'n thien va hinh dang dd thi cua cac ham sd lugng giac
neu tren.
2. KT nang
Sit dung thanh thao cdng thiic nghiem.
Giai thanh thao cac phucmg trinh lugng giac cof ban va mdt sd dang phucmg
trinh lugng giac khac.
Bie't xet su bie'n thien, ve dd thi cua cac ham sd lugng giac y — siiu,
y = cosx, y = tan X, y = cot x va mgt sd ham sd lugng giac dofn gian khac.
Giai thanh thao cac phucmg trinh lugng giac ca ban.
Bie't each giai mgt sd dang phucmg trinh lugng giac khdng qua phiic tap cd
the quy dugc ve phucmg trinh bac nhit va bac hai dd'i vdi mdt ham sd
lucmg giac.
3. Thai do
Tu giac, tich cue, ddc lap va chu ddng phat hien cung nhu linh hdi kie'n
thiic trong qua trinh hoat ddng.
cdn than, chinh xac trong lap luan va tinh toan.
Cam nhan dugc thuc te' cua toan hge, nh^t la ddi vdi lugng giac.
III. CAU TAO C O A C H U O N G

Du kie'n thuc hien trong 17 tiet, phan phdi cu thi nhu sau :

6

§ 1. Cac ham sd lugng giac

(3 tie't)

Luyen tap

(1 tie't)

§2. Phucmg trinh lugng giac cor ban

(3 tiet)

Luyen tap

(2 tiet)


§3. Mdt sd dang phuomg trinh lugng giac dom gian

(4 tiet)

Luyen tap

(2 tiet)

On tap va kiem tra chuomg 1.

(2 tie't)

IV. NHQNG DIEU CAN LUU Y TRONG CHUONG
1) Trudc day, toan bd vin de lugng giac nam trong chucmg trinh Dai sd va
Giai tfch 11. Trong chuomg trinh mdi, phin md diu ve lugng giac da dugc
gidi thieu d chucmg cudi cua Dai sd 10, bao gdm cac van dd xay dung cac
khai niem ccf ban nhu gdc va cung lugng giac, cac gia tri lugmg giac cua

gdc (cung) lugng giac va mgt sd cdng thiic lugng giac. Lugmg giac ldp 11
la su ndi tie'p chucmg trinh lugmg giac ldp 10. Dac diem dd ddi hdi giao
vien phai luu y nhae lai hay ggi md cho hge sinh nhd lai cac kidn thiic d
ldp 10 cd lien quan den bai hoc de de dang tiep thu kien thiic mdi.
2) O ldp 10 chi ndi den cac gid tri lugng giac ciia gdc hay cung lugng giac a.
Sang ldp 11, khi ndi den cac ham sd lugng giac y =sinx, y = cosx,
y = tan X, y = cotx ta hieu x la sd thuc va la sd do radian cua gdc hay
cung lugng giac.
3) Day la lan dau tien hge sinh lam quen vdi ham sd tuin hoan. Tuin hoan la
tfnh cha^t ndi bat cua cac ham sd lugng giac nen mac dii chuomg trinh
khdng yeu cau trinh bay tdng quat ve ham sd tuin hoan, cac tac gia vSn
gidi thieu dinh nghia ham sd tuin hoan (cudi §1) nham nhdc nhd hge sinh
chii y tfnh chat tuan hoan cua cac ham sd lugng giac.
4) Yeu ciu ve giai cac phucmg trinh lugng giac d day dugc giam nhe ra't nhidu
so vdi trudc day. Dieu dd the hien d hai didm co ban :
- Chi neu cac dang phuomg trinh dom gian, khdng ddi hdi phai cd nhiing
thu thuat bie'n ddi lugng giac phiic tap, va neu cd cac didu kien kem theo
thi viec thir lai cac dieu kien dd kha dem gian.
- Khdng yeu ciu giai va bien luan phucmg trinh lugng giac chiia tham sd.
Tuy nhien, giao vien can chu y ren luyen cho hoc sinh kT nang giai cac
phucmg trinh lugng giac ccr ban that thanh thao. Dd la ccf sd dd hoc sinh nang
cao kT nang giai cac phuomg trinh phiic tap hom.


P h a n 2>
CAC B A I SOAI^

§1. Cac ham so' Itfctog giac
(tiet 1, 2, 3)
I. MUCTIEU

1. Kie'n thurc
HS nam dugc :
Nhd lai bang gia tri lugng giac.
Ham sd y = sinx, ham sd y = cosx; su bidn thien, tfnh tuan hoan va cac tfnh
chat cua hai ham sd nay.
Ham sd y = tanx, ham sd y = cotx; su bidn thien, tfnh tuin hoan va cac tfnh
cha^t cua hai ham sd nay.
Tim hieu tfnh chit iuin hoan cua cac ham sd lugng giac.
Dd thi cua cac ham sd lugng giac.
2. KT nang
' • Sau khi hge xong bai nay, HS phai didn ta dugc tinh tuan hoan, chu ki tuin
hoan va su bidn thien ciia cac ham sd lugng giac.
Bieu didn dugc dd thi cua cac ham sd lugng giac.
Mdi quan he giiia cac ham sd y = sinx va y = cosx.
Md'i quan he giiia cac ham sd y = tanx va y = cotx.
3. Thai do
Tu giac, tfch cue trong hge tap.
Biet phan biet rd cac khai niem eg ban va van dung trong tiing trudng hgp
cu the.
Tu duy cac van de cua toan hge mgt each Idgic va he thdng.


II. CHUAN Bj COA GV VA HS
1. Chuan bj cua GV
Chudn bi cae cau hdi ggi md.
Chudn bi cac hinh tur hinh 1.1 de'n 1.13.
Chuan bi pha'n mau, va mdt sd dd dung khac.
2. Chuan bj cua HS
Cin dn lai mdt sd kidn thiic da hge ve lugng giac d ldp 10.
III. PHAN PHOI THOI LUONG

Bai nay chia lam 3 tie't r
Tie't 1 : Tif ddu den hit phdn I.
Tiet 2 : Tiep theo den hit phdn 2.
Tiet 3 : Tiep theo den hit phdn 3 vd bdi tap.
IV. TIEN TRINH DAY HOC
A. DAT VXN"DE

Cdu hoi I
Xet tinh diing - sai cua cac cau sau day :
a) Ndu a > b thi sina > sinb.
b) Ndu a > b thi cosa > cosb.
GV : Ca hai khang dinh tren deu sai. Cd the dan ra cae vf du eu the.
Cdu hoi 2
Nhiing cau sau day, cau nao khdng cd tfnh diing sai?
a) Neu a > b thi tana > tanb.
b) Neu a > b thi cota > cotb.
GV : Ta tha'y : Ca hai cau tren deu diing. Sau day, chiing ta se nghien ciiu ve cac
tfnh chat ciia cae ham sd y = sinx, y = cosx, y = tanx va y = cotx; su bie'n thien
va tfnh tudn hoan eiia eac ham sd dd.


B. BAI M6I
HOAT DONG 1
I, Cac ham sd y = sinx va y = cosx
• Thuc hien |H1| trong 3'
Muc dich.
Nhae lai each xac dinh sin x, cos x de chuydn tidp sang dinh nghTa cac ham
sd sin va cosin.
Hoqt dgng cua GV

Hoqt dpng cua HS
Gai y tra Idi cau hoi 1

Cdu hoi 1
Chi ra doan thang cd do dai dai
sd bang sinx
Cdu hoi 2

OK = sin X.

Ggi y tra ldi cau hoi 2

Chi ra doan thang cd do dai dai
sd bang cosx

OH = cosx.

GV: ggi hai HS trd ldi

•i:

Ggi y tra Icri cau hoi 3

Cdu hoi 3
.

It

Tinhsm—. cos
2

.

,COS27t.

4,

sm—= 1,
2
•

•^

1

( n] ^
0
1
cos — = — ,cos27t = l .

I 4j 2

a) Dinh nghia
• GV ggi hai hge sinh nhae lai cac gia tri lugng giac sin va cdsin. Sau dd GV
neu dinh nghia.
Quy tdc dgt tuong img mdi sdthitc x vdi sin cua gdc lugng giac cd sd
do radian bang x dugc ggi la hdm sd sin, ki hieu /ay = sinx .
Quy tdc dgt tucmg Ung mdi sdthuc x vdi cdsin ciia gdc lugng giac cd
sddo radian bdng x dugc ggi la hdm sd cosin, ki hieu Id y = cos x.
10

• GV neu cau hdi:
?1| So sanh sinx va sin(-x).
• GV neu nhan xet:
Ham sd y = sinx la mgt hdm sole vi sin(-x) = -sinx vdi mgi x thudc
• Thuc hien |H2| trong 3',
Muc dich. On lai dinh nghia ham sd chan.
Hoqt dong cua GV
Cdu hoi 1
So sanh cos a va cos(-a).
Cdu hoi 2

Hoqt dong cUa HS
Ggi y tra Idi cau hoi 1
Hai gia tri nay bang nhau.
Ggi y tra Icri cau hoi 2

Tai sao cd the khang dinh ham Ham sd y = cosx la mdt ham sd chan
sd y = cos\ la mdt ham sd chan? vi vdi mgi x e R ta cd
cos(-x) = cosx.

b) Tinh chdt tudn hodn cua hdm sdy = sinx vdy = cosx
• GV neu mdt so cau hdi nhu sau :
?2| So sanh : sin(x + ^27t) va sinx.
• Neu dinh nghla trong SGK.
Cac ham so y = sinx va y = cosx tudn hoan vdi chu ki 27c.
• GV dua ra tfnh chat:
Tii tfnh chdt tudp hoan vdi chu ki 2n, ta thay khi biet gia tri cac ham sd
y = sinx va y = cosx tren mdt doan cd do dai 2n (chang han doan [0; 2ii] hay
doan [-Tt; Tt]) thi ta tfnh duge gia tri cua chung tai mgi x.

c) Su bien thien ctia hdm sdy = sinx
• GV dua ra cau hdi

11


?3|

Neu lai chu ki tudn hoan cua ham sd y = sinx. Tfnh tudn hoan cua cae
ham sd dd cd lgi fch gi trong viec xet chidu bi6i thien cua cac ham
sd dd.

?4|

De xet chieu bidn thien eua cac ham sd dd ta can xet trong mdt
khoang cd do dai bao nhieu?

?5|

Hay neu mdt khoang de xet ma em eho la thuan lgi nha't.

• Sir dung cac hinh 1.2, 1.3 de md ta chidu bie'n thien cua ham sd dd trong doan
[-Tt; Tt].

,

B

0
A'\

^ - ^

1 J

r'

B'

?6

Tt .

Trong doan [-Tt;—] cac ham sd y = sinx ddng bidn hay nghich
bie'n?
Tt

21

Trong doan [ — ; 0] cac ham sd y = sinx ddng bidn hay nghich bidn?

73.

Trong doan [0; — ] cac ham sd y = sinx ddng bidn hay nghich bidn?

?9

Trong doan [—; Tt] cac ham sd y = sinx ddng bidn hay nghich bidn?
Sau khi cho hge sinh tra ldi, GV kdt luan va neu bang bidn thien
Tt

y = sinx
12

Tt

-Tt

2"

0

1

Tt


• De ve dd thi ham sd GV cdn eho HS tim mdt sd eac gia tri dac biet bang each
cho HS didn va chd trdng sau day :
X

0

y = sinx

...

Tt

Tt

Tt

Tt

2Tt

3Tt

5Tt

6

4

3

2

3

4

6

Tt

...

• GV sir dung hinh 1.5 va hinh 1.6 de neu dd thi cua ham sd tren.

• GV neu nhan xet trong SGK :
1) Khi X thay ddi, ham sd y = sinx nhan mgi gia tri thudc doan [-1; 1]. Ta ndi
tap gid tri ciia ham sd y = sinx la doan [-1; 1].
2) Ham sd y = sinx ddng bidn tren khoang

Tt

Tt

•2'2

Tuf dd, do tfnh chtft

tudn hoan vdi chu ki 2TI, ham sd y = sinx ddng bidn tren mdi khoang
- - + k2Tt; - + k2Tt \,

keZ.

• Thuc hien |H3| trong 3'
Muc dich
, (n 3n^ "
- Nhan bifit tfnh nghich bidn cua ham sd y = sinx tren khoang — ; — nhd
v2 2 /
dd thi (bang bidn thien chi mdi xet tren (-Tt; Tt)); dieu dd cdn giiip ren luyen
ki nang dgc.
- Nhd tfnh chdt tudn hoan vdi qhu ki 2Tt ciia ham sd y = sinx de suy ra ham sd
dd nghich bidn tren cac khoang — + k2Tt;
v2
2
Hoqt dpng cua GV

Cdu hoi 1

h k2Tt .
/
Hoqt dpng cua HS

Ggi y tra Idi cau hoi 1

Quan sat dd thi, ta thay ham sd
^Tt- 3n^ . .
Trong khoang — ; — ham so
U 2j
y = sinx nghich bidn
y = sinx ddng bien hay nghich , . , fTt 3Tt^
bidn?
khoang^-;-J

tren

13


Hoqt dpng cUa GV
Cdu hoi 2

Hoqt dpng cua HS
Ggi y tra loi c^u hoi 2

Ham sd y = nghich bidn tren Do tfnh chat tudn hoan vdi chu ki 2Tt,
mdi khoang

nd
nghich
bidn
tren
mgi
^Tt , ,,
3Tt , ^ '
—+ k2Tt; — + k2Tt , k e Z .

U

2

khoang —+ k2Tt ; — + k2Tt •)

U

j

2

J

k &Z.

d) Su bien thien cda hdm sdy = cosx
• GV dua ra cau hdi
?10| Neu lai chu ki tudn hoan cua ham sd y = cosx. Tfnh tudn hoan ciia
ham sd dd cd lgi fch gi trong viec xet chieu bien thien cua cac ham sd dd.
?11|

De xet chidu bien thien ciia ham sd dd ta cdn xet trong mdt khoang cd
dd dai bao nhieu?

?12| Hay neu m()t khoang dd xet ma em cho la thuan lgi nhdt.
Sii dung hinh 1. 8 de md ta chieu bien thien eua ham sd dd trong doan
[-Tt; Tt].
Tt .

?13| Trong doan [-Tt; — ] cac ham so y = cosx ddng bidn hay nghich bidn?
Tt

?14| Trong doan [ — ; 0] cac ham sd y = cosx ddng bidn hay nghich bien?
Tt

?15| Trong doan [0; — ] cac ham sd y = cosx ddng bidn hay nghich bidn?
Tt

?16| Trong doan [ —; Tt] cac ham sd y = cosx ddng bidn hay nghich bidn?
Sau khi cho hge sinh tra ldi GV kdt luan va neu bang bien thien
-Tt.

y = cosx
14

-1

0

Tt

-1


De ve dd thi ham sd GV cdn cho HS tim mdt so cac gia tri dac biet bang each
cho HS dien va chd trdng sau day :
0

Tt

Tt

Tt

Tt

2Tt

3Tt

'e

'4

3^

i

T

T

5Tt

T

Tt

y = cosx
GV sit dung hinh 1.7 de neu dd thi cua ham sd tren.
• Thuc hien .H4| trong 3'
Muc dich
Khao sat su bidn thien ciia ham sd y = cosx tren [-Tt; Tt] bang each quan sat
chuyen ddng cua hinh chieu H ciia didm M tren true cdsin (bd sung cho each
quan sat dd thi).
Hoqt ddng cua GV
Cdu hoi 1

Hoqt dpng cua HS
Ggi y tra Idi cau hoi I

Nhan xet vd tfnh tang, giam ciia Khi M chay tren dudng trdn lugng
ham sd y = cosx khi M chay tCt
giac theo chidu duomg tur A' den A,
A' ddn A.
hinh chidu H ciia M tren true cdsin
chay dgc true dd tir A' den A ntn OH
tlie la cosx tang tut -1 ddn 1;
Cdu hoi 2

Ggi y tra Idi cau hoi 2

Nhan xet ve tfnh tang, giam ciia Khi M chay tren dudng trdn lugng
ham sd y = cosx khi M chay tur giac theo chieu duomg tiir A de'n A',
A ddn A'
diem H chay dgc true cdsin tii A den A'
nen OH tlie la cosx giam tur 1 ddn - 1 .
• GV neu nhan xet trong SGK :
1) Khi X thay ddi, ham sd y = cosx nhan mgi gia tri thudc doan [rl; 1]. Ta ndi
tap gid tri ciia ham sd y = cosx la doan [-1; 1].
2) Do ham sd y = cosx la ham sd chSn nen dd thi ciia ham so y = cosx nhan true
tung lam true ddi xiing.
15


3) Ham sd y = cosx ddng bien tren khoang (-Tt; 0). Tii dd do tfnh chat tudn hoan
vdi chu ki 2Tt, ham sd y = cosx ddng bien tren mdi khoang (-Tt + ^2Tt; ^2Tt),
A:e Z.
Thuc hien |H5| trong 3'.
Muc dich
Xet tfnh ddng bie'n va nghich bie'n eua ham sd y = cosx tren doan [-Tt; Tt].
Hoqt dpng da GV

Hoqt dpng cua HS

Cdu hoi I
Ggi y tra Idi c^u hoi 1
Nhan xet vd tfnh ddng bien va Quan sat dd thi, ta thay ham sd
nghich bidn cua ham so
y = cosx nghich bidn tren khoang

y = cosx tren khoang (0; Tt).
(0; Tt).
Cdu hoi 2
Ggi y tra Idi cau hoi 2
Nhan xet vd tfnh ddng bidn va
Do tinh chat tudn hoan vdi chu ki 2Tt,
nghich bidn cua ham sd :
nd nghich bidn tren mgi khoang
y = cosx tren khoang
(Ikn; Tt + 2^Tt), ^ G Z.
(k2Tt; Tt + k2Tt).
• Dd neu bang ghi nhd : GV yeu cdu HS khdng sit dung SGK va didn vao chd
trdng sau:
Ham sd y = sinx
- Cd tap xac dinh la

Ham sd y = cosx
;

- Cd tap xac dinh la ....;

- Cd tap gia tri la ...;

- C d tap gia tri la ...;

- La ham so ...;

- L a h a m sd...;

- La ham sd tudn hoan vdi chu - La ham sd tudn hoan vdi chu

ki...;
ki...;
- Ddng bie'n tren mdi khoang
va nghich
khoang...

bidn

- Ddng bien tren mdi khoang ...
va nghich bie'n tren mdi khoang

tren mdi

- Cd dd thi la mgt dudng hinh - Cd dd thi la mdt dudng hinh sin.
sin.

16


nOATD(DNG2
2. Cac ham sd y = tanx va y = cotx
a) Dinh nghia
• Neu dinh nghia trong SGK.
Quy tdc dgt tuong iing mdi sdx e 3)^ vdi sd'thuc tan x • sinx duoc
cosx
ggi la hdm sdtang, ki hieu la y = tan x.
• GV dua ra eau hdi
?17| Ham sd y = tan x khdng xac dinh tai nhiing didm nao?
.cosx
Quy tdc ddt tuang ifng mdi sdx e 3)2 vdi sd'thuc cotx ='

sinx
ggi la hdm sd cdtang, ki hieu la y = cot x.

duac

?18| Ham sd y = cotx khdng xac dinh tai nhiing didm nao?
• GV sir dung hinh 1.9 va dua ra cac cau hdi:
?19| Tren hinh 1.9 hay chi ra cac doan thang ed do dai dai sd cua tanx va
cotx.
• GV neu nhan xet trong SGK:
1) Ham sd y = tanx la mdt hdm so le vi ndu x e 9)j thi -x e 9)i va
tan(-x) = -tanx.
2) Ham sd y = cotx ciing la mgt hdm sd le vi ndii x e ^2 tW ~x^ ^ ®2 ^^
cot(-x) = -cote.
b) Tinh tudn hodn
• GV dua ra cac cau hdi:
?20

So sanh tana va tan (a + kTt).

?21

So sanh cota va cot (a + kTt).

? 22

Nhan xet ve tfnh tudn hoan ciia hai ham sd tren.
17

• GV dua ra kdt luan cudi ciing:
T = Tt la sd duomg nhd nha't thoa man
tan(x + T) = tanx vdi mgi x G 9)j,
va T = Tt cung la sd duomg nhd nhdt thoa man
cot(x + T) = cotx vdi mgi x G 9)2Ta ndi cac ham sd y = tanx va y = cotx la nhiing hdm sdtudn hodn vdi
chu ki Tt.
c) Subien thien vd do thi cua hdm sdy = tanx
• GV dua ra cac cau hdi sau:
Sir dung hinh 1. 10 de md ta chidu bidn thien cua ham sd dd trong khoang
(--•-)
?23| Trong khoang ( —Tt; 0) ham sd y = tanx ddng bidn hay nghich bidn?
Tt .

?24| Trong khoang (0; —) ham sd y = tanx ddng bidn hay nghich bidn?
Tt TT

GV ke't luan : Ham sd y = tanx ddng bidn trong mdi khoang ( — ; —).
• Thue hien |H6| trong 5'
Hoqt dpng cua GV

Hoqt dpng cua HS

Cdu hoi 1
Ggi y tra Idi c^u hdi 1 '
Tai sao cd thd khang dinh ham Ta da bidt, ham sd y = tanx ddng
sd y = tanx ddng bidn tren mdi
bidn tren khoang — ; — nen do
khoang
Tl

Tt

2

2

— + kTt; — + kTt

,keZ?

tinh chdt tudn hoan vdi chu ki TT, nd
ddng bidn tren mgi khoang
— + kTt; — + kTi \,k G

2

18

2


• GV neu va md ta dd thi cua ham sd y = tanx qua hinh 1.11 trong SGK.
• GV neu cac nhan xet quan trgng sau :
1) Khi X thay ddi, ham sd y = tanx nhan mgi gia tri thue. Ta ndi tap gid tri cua
ham sd y = tan x la M.
2) Vi ham so y = tanx la ham sd le nen dd thi ciia nd nhan gd'c toa do lam tam
dd'i xiing.
Tt

3) Ham sd y = tanx khdng xac dinh tai x = — + kTt (^ G Z). Vdi mdi k G Z,

Tt

dudng thang vudng gdc vdi true hoanh, di qu^ didm — + kTt; 0 ggi la mdt
2
dudng tiem can ciia dd thi ham sdy = tan x.
d) Subien thien cua hdm sdy = cotx
• GV dua ra cac cau hdi sau dd HS khao sat.
Tt

1

?25| Trong khoang (0; —) ham sd y = cotx ddng bie'n hay nghich bidn?
1

Tt

,

V

?24| Trong khoang (—; Tt) ham so y = cotx ddng bidn hay nghich bidn?
z
GV kdt luan : Ham sd y =; cotx ddng bidn trong mdi khoang (0; TI).
Sau dd GV sit dung hinh 1.12 de md ta dd thi cua ham sd y = cotx.
• Dd ghi nhd GV cho HS dien vao chd trdng sau:
Ham so y = tanx
- Cd tap xac dinh la ...;
- Cd tap gia tri la...;
- L a ham sd...;
- La ham so tudn hoan vdi ehu ki...;

- Ddng bidn tren mdi khoang ...

Ham sd y = cote
- Cd tap xac dinh la : ....
- Cd tap gia tri la ...;
-Laham sd...;
- La ham sd tudn hoan vdi chu
ki...;
- Nghich bidn tren mdi khoang

- Cd dd thi nhan mdi dudng thang - Cd dd thi nhan mdi dudng
... lam mdt dudng tiem can.
thang ... lam mdt dudng tiem
can.
19


HOAT DONG 3
2. Ve khai niem ham so tuan hoan
• GV neu khai niem ham sd tudn hoan:
Hdm sdy = i(x) xdc dinh tren tap hgp 3)dugc ggi Id hdm sdtudn
hodn niu cd sd'T ^0 sao cho vdi mgi x e 3) tacd
x + T e9),x-T

e3)vaf(x + T)^jix).

Niu cd sd'T duang nho nhdt thod mdn cdc dieu kien tren thi hdm so
dd dugc ggi la mdt hdm sdtudn hodn v&i chu ki T.
Sau dd GV dua ra mgt sd cau hdi nhdm nhdn manh vd ham tudn hoan va chu
ki cua ham sd tudn hoan.

?25| Ham sd y = 2sin x tudn hoan hay khdng? Ndu la ham so tudn hoan
hay chi ra chu ki?
?26| Ham sd y = -32cos x tudn hoan hay khdng? Ndu la ham sd tudn hoan
hay chi ra chu ki?
?27| Ham sd y = 2sin — tudn hoan hay khdng? Ndu la ham sd tudn hoan
hay chi ra ehu ki?
?27| Ham sd y = Stan x tudn hoan hay khdng? Ndu la ham sd tudn hoan
hay chi ra chu ki?
?28| Ham sd y = -3cot x tudn hoan hay khdng? Ndu la ham sd tudn hoan
hay chi ra chu ki?
?29| Ham sd y = 2cot2x tudn hoan hay khdng? Ndu la ham sd tudn hoan
hay chi ra chu ki?
Sau dd GV dua ra cac eau hdi sau nham cung cd bai hge:
Chon diing sai md em cho Id hop ly.
n.
?30| Ham sd y = sinx nghich bien tren khoang (0; —).
(a),Dung;
20

(b) Sai.


1

. 1

'

•>

/ ^

\

?31| Ham sd y = sinx ddng bien tren khoang (—; Tt).
(a) Diing;

(b) Sai.

1

Tt

?32| Ham sd y = sinx nghich bidn tren khoang (—; Tt).
(a) Diing;
1

(b) Sai.
V

Tt

?33| Ham so y = cosx ddng bidn tren khoang (—; 0).
(a) Dung;
1

(b) Sai.

^

Tt

?34| Ham so y = cosx nghich bien tren khoang (0; —).
(a) Dung;

(b) Sai.

1

Tt

?35| Ham sd y = cosx nghich bidn tren khoang (—; 0).
(a) Diing;
1

(b) Sai.
V

,

Tt

?36| Ham sd y = cosx ddng bidn tren khoang (0; —).
(a) Diing;
1

(b) Sai.
V

,

Tt

?37| Ham so y = tanx ddng bidn tren khoang (—; 0).
I

1

(a) Diing;

1

(b) Sai.
V

^

Tt
Tt

?39| Ham sd y = tanx nghich bidn tren khoang (—; 0).
?38| Ham sd y = tanx ddng bien tren khoang (0; —).
1

(a) Dung;
(a)
Diing;

(b)
(b) Sai.

Sai.

Tt

?40| Ham sd y = tanx nghich bidn tren khoang (0; —).
(a) Dung;

(b) Sai.
21


HOAT DONG 4

TOM TAT BAI H O C
1. Quy tac dat tucmg iing mdi sd thuc x vdi sd thuc y = sinx. Quy tac nay dugc
ggi la ham sd sin.
sin :

R -> M
X i-> y = sinx.

• y = sinx xac dinh vdi mgi x e R va -1 < sinx < 1.
• y = sinx la ham sd le.
• y = sinx la ham so tudn hoan vdi chu ki 2;r.

°^f

Ham sd y = sinx ddng bidn tren

va nghich bidn tren

Tt

2. Quy tac dat tucmg umg mdi sd thuc x vdi sd thuc y = cosx (h. 2b). Quy tdc nay
dugc ggi la ham sd cosin.
cosin : R -> R
X h^ y = cosx

• y = cosx xac dinh vdi mgi x G R va -1 < cosx < 1.
• y = cosx la ham sd chdn.
• y = cosx la ham sd tudn hoan vdi chu ki 2;r.
Ham sd y = cosx ddng bien tren doan [-Tt; 0] va nghich bidn tren doan [0; TT].
3. Ham sd tang la ham sd dugc xae dinh bdi cdng thiic
y = tanx =

sin X

(cosx it 0).

cosx

^-

Tap xac dinh eua ham sd y = tanx la 3)i = R \ {— + kTt I k G Z
)•

•

1 2
Tt

• y = tanx xac dinh vdi moi x 9^ — + ICTT, k e Z.
^

22

•

•

2


• y = tanx la ham sd le.
• y = tanx la ham sd tudn hoan vdi chu ki n.
Ham Sd y = tanx ddng bidn tren nura khoang

°1

4. Ham so cdtang la ham sd dugc xac dinh bdi cdng thiic
cosx

..
^.
(sinx -^ 0).

y = cot =

sinx
Tap xac dinh cua ham sdy = cotx la %2 - ^ ^ \kTi | it G Z}

• y = cotx la ham so tudn hoan vdi ehu ki it.
• y - cote la ham sd le.
vay ham sd y = cote nghich bidn tren khoang (0; ri).
5. Ham sd y = %x) xac dinh tren tap hgp 9) dugc ggi la ham so tudn hoan ne'u
cd sd r 9t 0 sao cho vdi mgi x

G

9) ta ed

x + rG9),x-rG9)vaXj<: + r ) = j(x).
Ndu cd sd T ducmg nhd nhdt thoa man cac didu kien tren thi ham sd dd dugc
ggi la mdt ham so' tuan hoan vdi chu ki T.
HOAT DONG 5
MOT SO CAU HOI T R A C NGHlfiM ON T A P
Cdu I.

BAI

1

(a) Tap xac dinh cua ham sd y = tan x la R.
(b) Tap xac dinh ciia ham sd y = cot x la ]
(c) Tap xac dinh cua ham sd y = cosx la''.
(d) Tap xac dinh cua ham sd y =

la'.
cosx

Trd ldi. (c).

23