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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