Thiết kế bài giảng đại số và giải tích nâng cao lớp 11 tập 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

NHA
XUAT
BAN
HA
NOI
Chiu
trdch
nhiem
xudt
bdn:
NGUYfiN
KHAC
OANH
Bien tap:
PHAM
QUOC
TUAN
Vebia:
NGUYfiN
TUXN
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
COA
CHUONG
Du kie'n thuc hien trong
17
tiet, phan phdi cu thi nhu sau :
§
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)
6
§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.
Phan

2>
CAC
BAI
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
Cdu hoi 1
Chi ra doan thang cd do dai dai
sd bang sinx
Cdu hoi 2
Chi ra doan thang cd do dai dai
sd bang
cosx
GV: ggi hai HS trd ldi
Cdu hoi 3
.
It
Tinhsm—. cos
2
.
4,
,COS27t.
Hoqt
dpng
cua HS
Gai y tra
Idi
cau hoi 1
OK
= sin

X.
Ggi y tra ldi cau hoi 2
OH =
cosx.
•i:
Ggi y tra Icri cau hoi 3
•
•^
1
sm—= 1,
2
( 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
Tai sao cd the khang dinh ham
sd y =
cos\
la mdt ham sd chan?
Hoqt dong
cUa
HS

Ggi y tra Idi cau hoi 1
Hai gia tri nay bang nhau.
Ggi y tra Icri cau hoi 2
Ham sd y = cosx la mdt ham
vi vdi mgi x
e R
ta cd
cos(-x) = cosx.
sd chan
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].
,
A'\
^-^
B
0
1
J
B'
r '
?6
Tt
.
Trong doan
[-Tt;—]
cac ham sd y = sinx ddng bidn hay nghich
bie'n?
21
73.
?9
Tt
Trong doan
[
—; 0] cac ham sd y = sinx ddng bidn hay nghich bidn?
Trong doan
[0;
— ]
cac ham sd y = sinx ddng bidn hay nghich bidn?
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
0
Tt
Tt
2"
1
Tt
y
= sinx
12
• 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
y = sinx
0

Tt
6
Tt
4
Tt
3
Tt

2
2Tt
3
3Tt
4

5Tt
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;
h k2Tt
.
v2 2
/
Hoqt
dpng
cua GV
Cdu hoi 1
Trong khoang
^Tt-
3n^
. .
—
; — ham so
U
2 j
y = sinx ddng bien hay nghich
bidn?
Hoqt dpng
cua HS
Ggi y tra Idi cau hoi 1
Quan sat dd thi, ta thay ham sd
y = sinx nghich bidn tren
,.
,
fTt
3Tt^

khoang^-;-J
13
Hoqt dpng
cUa
GV
Cdu hoi 2
Ham sd y = nghich bidn tren
mdi khoang
^Tt
,
,,
3Tt ,
^ '
—+
k2Tt;
—
+ k2Tt
U
2 j
,keZ.
Hoqt dpng cua HS
Ggi
y tra
loi
c^u
hoi 2
Do tfnh chat tudn hoan vdi chu ki 2Tt,
nd nghich bidn tren mgi
khoang
—+ k2Tt ; —+ k2Tt

U
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.
0
Tt

y = cosx
-1
-1
14
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 :
Tt
Tt Tt Tt 2Tt 3Tt 5Tt
'e
'4
3^
i
T
T T
0
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
Nhan xet vd tfnh tang, giam ciia
ham sd y = cosx khi M chay
tCt
A' ddn A.
Cdu hoi 2
Nhan xet ve tfnh tang, giam ciia
ham sd y = cosx khi M chay tur
A ddn A'
Hoqt dpng cua HS
Ggi y tra Idi cau hoi I
Khi M chay tren dudng trdn lugng
giac theo chidu duomg tur A' den 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;
Ggi y tra Idi cau hoi 2
Khi M chay tren dudng trdn lugng

giac theo chieu duomg tiir A de'n 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
Cdu
hoi
I
Nhan
xet vd
tfnh ddng bien
va
nghich bidn
cua
ham
so
y
=
cosx tren khoang (0; Tt).
Cdu
hoi
2
Nhan
xet vd
tfnh ddng bidn
va
nghich
bidn

cua ham sd :
y
=
cosx tren khoang
(k2Tt;
Tt
+
k2Tt).
Hoqt dpng cua
HS
Ggi
y tra Idi
c^u
hoi 1
Quan
sat dd
thi,
ta
thay ham
sd
y
=
cosx nghich bidn tren khoang
(0;
Tt).
Ggi
y tra Idi cau hoi 2
Do tinh chat tudn hoan
vdi chu ki
2Tt,

nd nghich bidn tren
mgi
khoang
(Ikn;
Tt
+
2^Tt),
^ G
Z.
•
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 ;
-
Cd
tap
gia tri la ;
-
La
ham
so
;

-
La ham sd
tudn hoan
vdi chu
ki ;
- Ddng bie'n tren
mdi
khoang
va nghich bidn tren
mdi
khoang
-
Cd dd thi la mgt
dudng hinh
sin.
Ham
sd
y
=
cosx
-
Cd tap xac
dinh
la ;
-Cd
tap gia tri la ;
-Laham
sd ;
-
La ham sd

tudn hoan
vdi chu
ki ;
- Ddng bien tren
mdi
khoang

va nghich bie'n tren mdi khoang
-
Cd dd thi la mdt
dudng hinh
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 •
ggi la hdm
sdtang,

ki
hieu
la y
=
tan x.
• GV dua ra eau hdi
sinx
cosx
duoc
?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
='
duac
sinx
ggi la hdm sd
cdtang,
ki
hieu
la y

=
cot x.
?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)2-
Ta 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
( •-)
Tt
?23|
Trong khoang (—; 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
Tai sao cd thd khang dinh ham
sd y
=
tanx ddng bidn tren mdi
khoang
Tl
Tt
— +
kTt;
— +
kTt
2 2
,keZ?
Ggi y tra Idi
c^u
hdi
1'
Ta da bidt, ham sd y = tanx ddng

bidn tren khoang — ;
—
nen do
tinh chdt tudn hoan vdi chu
ki
TT,
nd
ddng bidn tren mgi khoang
—
+
kTt;
—+
kTi
\,k
G
2 2
18
•
GV
cua
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
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,
dudng thang vudng gdc vdi
true
hoanh, di
qu^
didm
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
—
+
kTt;
0
2
ggi la mdt
1
Tt
?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 ;
-La ham
sd ;
- La ham so tudn hoan vdi ehu
ki ;
- Ddng bidn tren mdi khoang
- Cd dd thi nhan mdi dudng thang
lam mdt dudng tiem can.
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 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;
(b) Sai.
20
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; (b) Sai.
1
V
Tt
?33|
Ham so y = cosx ddng bidn tren khoang (—; 0).
(a) Dung; (b) Sai.
1
^
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; (b) Sai.
1
V
,
Tt

?36|
Ham sd y = cosx ddng bidn tren khoang
(0;
—).
(a) Diing; (b) Sai.
1
V
,
Tt
?37|
Ham so y = tanx ddng bidn tren khoang (—; 0).
(a) Diing; (b) Sai.
1
V
^
Tt
?38|
Ham sd y = tanx ddng bien tren khoang
(0;
—).
(a) Diing; (b) Sai.
I
1
Tt
?39|
Ham sd y = tanx nghich bidn tren khoang (—; 0).
(a) Dung; (b) Sai.
1
Tt
?40|

Ham sd y = tanx nghich bidn tren khoang
(0;
—).
(a) Dung; (b) Sai.
21
HOAT DONG
4
TOM
TAT
BAI
HOC
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.
Ham sd
y
= sinx ddng bidn tren
°^f
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
sin X
y = tanx
=
(cosx
it
0).
cosx
^
-
Tap xac dinh eua ham sd
y
= tanx la
3)i =
R \
{—
+ kTt
I
k
G

Z
)•
•
12
Tt
•
y
=
tanx xac dinh vdi moi x
9^
—
+
ICTT,
k e
Z.
^ • •
2
22
•
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
4.
Ham so cdtang la ham sd dugc xac dinh bdi cdng thiic
°1

cosx
.
.
^.
y = cot
= (sinx -^
0).
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
TRAC
NGHlfiM ON
TAP
BAI
1
Cdu
I.
(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