HÀM SỐ LƯỢNG GIÁC

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H?M S LNG GI?C

A. T?M TT L? THUYT

I. C?c c?ng thc lng gi?c

1. C?c hng ng thc:

* sin2 + cos2 = 1 vi mi

* tan .cot = 1

vi mi k 2

*

1+

tan2

=

1 cos2

vi mi k2

*

1+

cot2

=

1 sin2

vi mi k

2. H thc c?c cung c bit

a.Hai cung i nhau: v? -

cos(-) = cos

sin(-) = -sin

tan(-) = - tan

cot(-) = -cot

b. Hai cung ph nhau: v? - 2

cos( - ) = sin 2

tan( - ) = cot 2

c. Hai cung b? nhau: v? - sin( - ) = sin

sin( - ) = cos 2

cot( - ) = tan 2

cos( - ) = - cos

tan( - ) = - tan

cot( - ) = -cot

d) Hai cung hn k?m nhau : v? + sin( + ) = -sin

cos( + ) = - cos

tan( + ) = tan

cot( + ) = cot

3. C?c c?ng thc lng gi?c a. C?ng thc cng cos(a b) = cosa.cos b sina.sin b

sin(a b) = sina.cos b cosa.sin b

tan(a b) = tan a tan b 1 tan a.tan b

b) C?ng thc nh?n

sin 2a = 2 sina cosa

cos 2a = cos2 a - sin2 a = 1 - 2sin2 a = 2cos2 a - 1

sin 3a = 3sina - 4sin3 a

cos3a = 4cos3 a - 3cosa

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c. C?ng thc h bc

sin2 a = 1 - cos 2a 2

cos2 a = 1 + cos 2a 2

tan2 a = 1 - cos 2a 1 + cos 2a

d. C?ng thc bin i t?ch th?nh tng

cosa.cos b = 1[cos(a - b) + cos(a + b)] 2

sina.sin b = 1 [cos(a - b) - cos(a + b)] 2

sina.cos b = 1[sin(a - b) + sin(a + b)] . 2

e. C?ng thc bin i tng th?nh t?ch

cosa + cos b = 2cos a + b .cos a - b

2

2

cosa - cos b = -2sin a + b .sin a - b

2

2

sina + sin b = 2sin a + b .cos a - b

2

2

sina - sin b = 2cos a + b .sin a - b

2

2

tana + tan b = sin(a + b) cosa cos b

tana - tan b = sin(a - b) . cosa cos b

II. T?nh tun ho?n ca h?m s

nh ngha: H?m s y = f(x) x?c nh tr?n tp D c gi l? h?m s tun

ho?n nu c? s T 0 sao cho vi mi x D ta c? x T D v? f(x + T) = f(x) .

Nu c? s T dng nh nht tha m?n c?c iu kin tr?n th? h?m s ? c gi l? h?m s tun ho?n vi chu k? T . III. C?c h?m s lng gi?c 1. H?m s y = sin x

? Tp x?c nh: D = R ? Tp gi?c tr: [ - 1;1] , tc l? -1 sin x 1 x R

? H?m s ng bin tr?n mi khong (- + k2; + k2) , nghch bin tr?n

2

2

mi khong ( + k2; 3 + k2) .

2

2

? H?m s y = sin x l? h?m s l n?n th h?m s nhn gc ta O l?m

t?m i xng. ? H?m s y = sin x l? h?m s tun ho?n vi chu k? T = 2 .

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? th h?m s y = sin x .

y

-5

-

2

-2

- 2

-3

-3

2

1 O

2

2

3

2 2

3x 5 2

2. H?m s y = cos x ? Tp x?c nh: D = R ? Tp gi?c tr: [ - 1;1] , tc l? -1 cos x 1 x R ? H?m s y = cos x nghch bin tr?n mi khong (k2; + k2) , ng bin tr?n mi khong (- + k2; k2) . ? H?m s y = cos x l? h?m s chn n?n th h?m s nhn trc Oy l?m trc i xng. ? H?m s y = cos x l? h?m s tun ho?n vi chu k? T = 2 . ? th h?m s y = cos x . th h?m s y = cos x bng c?ch tnh tin th h?m s y = sin x theo v?c t v = (- ; 0) .

2

y

-5

-

2

-2

- 2

-3

-3

2

1

O 2

3

2 2

3x 5 2

3. H?m s y = tanx

? Tp x?c nh : D =

\

2

+

k,

k

? Tp gi? tr: ? L? h?m s l ? L? h?m s tun ho?n vi chu k? T =

?

H?m

ng

bin

tr?n

mi

khong

-

2

+

k;

2

+

k

? th nhn mi ng thng x = + k, k l?m mt ng tim cn. 2

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? th

y

-

-2

-

2

2

-5

-3

O

2

2

3

5

2 2

2x

4. H?m s y = cot x

? Tp x?c nh : D = \k, k

? Tp gi? tr: ? L? h?m s l ? L? h?m s tun ho?n vi chu k? T =

? H?m nghch bin tr?n mi khong (k; + k)

? th nhn mi ng thng x = k, k l?m mt ng tim cn. ? th

y

-

-2

-

2

2

-5

-3

O

2

2

3

5

2 2

2x

B.PHNG PH?P GII TO?N.

Vn 1. Tp x?c nh v? tp gi? tr ca h?m s

Phng ph?p . ? H?m s y = f(x) c? ngha f(x) 0 v? f(x) tn ti ? H?m s y = 1 c? ngha f(x) 0 v? f(x) tn ti.

f(x)

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? sin u(x) 0 u(x) k, k

? cos u(x) 0 u(x) + k, k . 2

? -1 sinx, cosx 1. C?c v? d

V? d 1. T?m tp x?c nh ca h?m s sau: 1. y = tan(x - )

6

2. y = cot2( 2 - 3x) 3

Li gii.

1. iu kin: cos(x - ) 0 x - + k x 2 + k

6

62

3

TX: D =

\

2 3

+

k,

k

.

2. iu kin: sin( 2 - 3x) 0 2 - 3x k x 2 - k

3

3

93

TX: D =

\

2 9

+

k

3

,

k

.

V? d 2. T?m tp x?c nh ca h?m s sau:

1. y = tan 2x + cot(3x + )

sin x + 1

6

2. y = tan 5x sin 4x - cos 3x

Li gii.

1.

iu

kin:

sin x -1

sin(3x

+

) 6

0

x x

- + k2 2

- + k 18 3

Vy TX: D =

\ -

2

+

k2,

-

18

+

n 3

;

k,

n

2.

Ta

c?:

sin 4x

-

cos 3x

=

sin 4x

-

sin

2

-

3x

=

2

cos

x 2

+

4

sin

7x 2

-

4

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cos 5x 0

iu

kin:

cos

x 2

+

4

0

x

10

+

k

5

x

2

+

k2

sin

7x 2

+

4

0

x

-

14

+

k2 7

Vy TX: D =

\

10

+

k 5

,

2

+

n2,

-

14

+

2m 7

.

C?C B?I TO?N LUYN TP B?i 1 T?m tp x?c nh ca h?m s sau:

1. y = 1 - sin 2x cos 3x - 1

3. y = tan(2x - ) 4

2. y = 1 - cos 3x 1 + sin 4x

4. y = 1 + cot2 x 1 - sin 3x

B?i 2 T?m tp x?c nh ca h?m s sau:

1. y =

1

sin 2x - cos 3x

2. y =

tan 2x

3 sin 2x - cos 2x

3. y = cot x 2 sin x - 1

4. y = tan(x - ).cot(x - )

4

3

B?i 3 T?m tp x?c nh ca h?m s sau:

1. y = tan(2x + ) 3

3. y =

2 + sin x tan2 x

5. y = sin 3x sin 8x - sin 5x

2. y = tan 3x.cot 5x

4. y = tan 3x + cot(x + ) 3

6. y = tan 4x cos 4x + sin 3x

Vn 2. T?nh cht ca h?m s v? th h?m s

Phng ph?p . Cho h?m s y = f(x) tun ho?n vi chu k? T * kho s?t s bin thi?n v? v th ca h?m s, ta ch cn kho s?t v? v th h?m s tr?n mt on c? d?i bng T sau ? ta tnh tin theo c?c v?c

t k.v (vi v = (T; 0), k ) ta c to?n b th ca h?m s. * S nghim ca phng tr?nh f(x) = k , (vi k l? hng s) ch?nh bng s giao im ca hai th y = f(x) v? y = k .

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* Nghim ca bt phng tr?nh f(x) 0 l? min x m? th h?m s y = f(x)

nm tr?n trc Ox . Ch? ?: ? H?m s f(x) = a sin ux + bcos vx + c ( vi u,v ) l? h?m s tun ho?n vi

chu k? T = 2 ( (u,v) l? c chung ln nht). (u, v)

? H?m s f(x) = a.tan ux + b.cot vx + c (vi u,v ) l? h?m tun ho?n vi

chu k? T = . (u, v)

C?c v? d

V? d 1. X?t t?nh tun ho?n v? t?m chu k? c s ca c?c h?m s : f(x) = cos 3x .cos x

22

Li gii.

Ta c?

f(x) = 1 (cos x + cos 2x)

2

h?m s tun ho?n vi chu k? c s

T0

= 2 .

V? d 2. X?t t?nh tun ho?n v? t?m chu k? c s (nu c?) ca c?c h?m s sau.

( ) 1. f(x) = cos x + cos 3.x

2. f(x) = sin x2

Li gii. 1. Gi s h?m s ? cho tun ho?n c? s thc dng T tha

f(x + T) = f(x) cos(x + T) + cos 3(x + T) = cos x + cos 3x

Cho x = 0 cos T + cos

3T

=

2

cos cos

T=1 3T =

1

T

=

2n

3T = 2m

3 = m v? l?, do m,n n

m l? s hu t. n

Vy h?m s ? cho kh?ng tun ho?n.

2. Gi s h?m s ? cho l? h?m s tun ho?n

T 0 : f(x + T) = f(x) sin(x + T)2 = sin x2 x

Cho x = 0 sinT2 = 0 T2 = k T = k

f(x + k) = f(x) x .

( )2

Cho x = 2k ta c?: f( 2k) = sin k2 = sin(k2) = 0 .

( ) ( ) 2

f(x + k) = sin k2 + k = sin 3k + 2k 2 = sin(2k 2)

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f(x + k) 0 . Vy h?m s ? cho kh?ng phi l? h?m s tun ho?n.

V? d 3. Cho a,b,c,d l? c?c s thc kh?c 0. Chng minh rng h?m s f(x) = a sin cx + bcosdx l? h?m s tun ho?n khi v? ch khi c l? s hu t.

d Li gii. * Gi s f(x) l? h?m s tun ho?n T 0 : f(x + T) = f(x) x

a sin cT + b cosdT = b cosdT = 1 Cho x = 0,x = -T -a sin cT + b cosdT = b sin cT = 0

dT cT

= 2n = m

c d

=

m 2n

.

* Gi s c k,l : c = k . t T = 2k = 2l

d

dl

cd

Ta c?: f(x + T) = f(x) x f(x) l? h?m s tun ho?n vi chu k?

T = 2k = 2l . cd

V? d 4. Cho h?m s y = f(x) v? y = g(x) l? hai h?m s tun ho?n vi chu k

ln lt l? T1 ,T2 . Chng minh rng nu

T1 T 2

l? s hu t th? c?c h?m s

f(x) g(x); f(x).g(x) l? nhng h?m s tun ho?n.

Li gii.

V? T1 l? s hu t n?n tn ti hai s nguy?n m,n; n 0 sao cho T 2

T1 T2

=

m n

nT1

= mT2

= T

Khi ? f(x + T) = f(x + nT1) = f(x) v? g(x + T) = g(x + mT2 ) = g(x)

Suy ra f(x + T) g(x + T) = f(x) g(x) v? f(x + T).g(x + T) = f(x).g(x) ,

f(x + T) = f(x) . T ? ta c? iu phi chng minh. g(x + T) g(x)

Nhn x?t: 1. H?m s f(x) = a sin ux + bcos vx + c ( vi u,v chu k? T = 2 ( (u,v) l? c chung ln nht).

(u, v)

) l? h?m s tun ho?n vi

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