ds10-c6a.pdf
TRANSCRIPT
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Lng gic Trn S Tng
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I. Gi tr lng gic ca gc (cung) lng gic 1. nh ngha cc gi tr lng gic Cho OA OM( , ) = . Gi s M x y( ; ) .
( )
x OHy OK
AT k
BS k
cossin
sintancos 2coscotsin
= == =
= = +
= =
Nhn xt: , 1 cos 1; 1 sin 1
tan xc nh khi k k Z,2
+ cot xc nh khi k k Z,
ksin( 2 ) sin + = ktan( ) tan + =
kcos( 2 ) cos + = kcot( ) cot + =
2. Du ca cc gi tr lng gic
3. Gi tr lng gic ca cc gc c bit
0 6
4
3
2 2
3 3
4 3
2 2
00 300 450 600 900 1200 1350 1800 2700 3600
sin 0 12
22
32
1 32
22
0 1 0
cos 1 32
22
12
0 12
22
1 0 1
tan 0 33
1 3 3 1 0 0
cot 3 1 3
3 0 3
3 1 0
CHNG VI GC CUNG LNG GIC CNG THC LNG GIC
Phn t Gi tr lng gic I II III IV
cos + + sin + + tan + + cot + +
cosin O
cotang
s
in
tang
H A
M K B S
T
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Trn S Tng Lng gic
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4. H thc c bn: 2 2sin cos 1 + = ; tan .cot 1 = ; 2 2
2 21 11 tan ; 1 cot
cos sin
+ = + =
5. Gi tr lng gic ca cc gc c lin quan c bit
II. Cng thc lng gic 1. Cng thc cng
2. Cng thc nhn i sin 2 2sin .cos = 2 2 2 2cos2 cos sin 2 cos 1 1 2sin = = =
2
22 tan cot 1tan 2 ; cot 2
2 cot1 tan
= =
sin( ) sin .cos sin .cosa b a b b a+ = + sin( ) sin .cos sin .cosa b a b b a = cos( ) cos .cos sin .sina b a b a b+ = cos( ) cos .cos sin .sina b a b a b = +
tan tantan( )1 tan .tan
a ba b
a b+
+ =
tan tantan( )1 tan .tan
a ba b
a b
=+
H qu: 1 tan 1 tantan , tan4 1 tan 4 1 tan
+ + = = +
Gc hn km Gc hn km 2
sin( ) sin + = sin cos2
+ =
cos( ) cos + = cos sin2
+ =
tan( ) tan + = tan cot2
+ =
cot( ) cot + = cot tan2
+ =
Gc i nhau Gc b nhau Gc ph nhau
cos( ) cos = sin( ) sin = sin cos2
=
sin( ) sin = cos( ) cos = cos sin2
=
tan( ) tan = tan( ) tan = tan cot2
=
cot( ) cot = cot( ) cot = cot tan2
=
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Lng gic Trn S Tng
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3. Cng thc bin i tng thnh tch
4. Cng thc bin i tch thnh tng
Cng thc h bc Cng thc nhn ba (*) 2
2
2
1 cos2sin2
1 cos2cos2
1 cos2tan1 cos2
=
+=
=
+
3
3
3
2
sin 3 3sin 4sincos3 4 cos 3cos
3tan tantan 31 3 tan
= =
=
cos cos 2 cos .cos2 2
a b a ba b
+ + =
cos cos 2sin .sin2 2
a b a ba b
+ =
sin sin 2sin .cos2 2
a b a ba b
+ + =
sin sin 2 cos .sin2 2
a b a ba b
+ =
sin( )tan tancos .cos
a ba b
a b+
+ =
sin( )tan tancos .cos
a ba b
a b
=
sin( )cot cotsin .sin
a ba b
a b+
+ =
b aa ba b
sin( )cot cotsin .sin
=
sin cos 2.sin 2.cos4 4
+ = + =
sin cos 2 sin 2 cos4 4
= = +
1cos .cos cos( ) cos( )21sin .sin cos( ) cos( )21sin .cos sin( ) sin( )2
a b a b a b
a b a b a b
a b a b a b
= + +
= +
= + +
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Trn S Tng Lng gic
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VN 1: Du ca cc gi tr lng gic xc nh du ca cc gi tr lng gic ca mt cung (gc) ta xc nh im nhn
ca cung (tia cui ca gc) thuc gc phn t no v p dng bng xt du cc GTLG. Bi 1. Xc nh du ca cc biu thc sau:
a) A = 0 0sin 50 .cos( 300 ) b) B = 0 21sin 215 .tan7
c) C = 3 2cot .sin5 3
d) D = c 4 4 9os .sin .tan .cot5 3 3 5
Bi 2. Cho 0 00 90< < . Xt du ca cc biu thc sau: a) A = 0sin( 90 ) + b) B = 0cos( 45 )
c) C = 0cos(270 ) d) D = 0cos(2 90 ) +
Bi 3. Cho 02
< < . Xt du ca cc biu thc sau:
a) A = cos( ) + b) B = tan( )
c) C = 2sin5
+
d) D = 3cos8
Bi 4. Cho tam gic ABC. Xt du ca cc biu thc sau: a) A = A B Csin sin sin+ + b) B = A B Csin .sin .sin
c) C = A B Ccos .cos .cos2 2 2
d) D = A B Ctan tan tan2 2 2
+ +
Bi 5. a)
VN 2: Tnh cc gi tr lng gic ca mt gc (cung) Ta s dng cc h thc lin quan gia cc gi tr lng gic ca mt gc, t gi tr
lng gic bit suy ra cc gi tr lng gic cha bit. I. Cho bit mt GTLG, tnh cc GTLG cn li 1. Cho bit sin, tnh cos, tan, cot
T 2 2sin cos 1 + = 2cos 1 sin = .
Nu thuc gc phn t I hoc IV th 2cos 1 sin = .
Nu thuc gc phn t II hoc III th 2cos 1 sin = .
Tnh sintancos
= ; 1cot
tan
= .
2. Cho bit cos, tnh sin, tan, cot
T 2 2sin cos 1 + = 2sin 1 cos = .
Nu thuc gc phn t I hoc II th 2sin 1 cos = .
Nu thuc gc phn t III hoc IV th 2sin 1 cos = .
Tnh sintancos
= ; 1cot
tan
= .
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Lng gic Trn S Tng
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3. Cho bit tan, tnh sin, cos, cot
Tnh 1cottan
= .
T 22
1 1 tancos
= + 2
1cos1 tan
= +
.
Nu thuc gc phn t I hoc IV th 2
1cos1 tan
=+
.
Nu thuc gc phn t II hoc III th 2
1cos1 tan
= +
.
Tnh sin tan .cos = . 4. Cho bit cot, tnh sin, cos, tan
Tnh 1tancot
= .
T 221 1 cot
sin
= +
2
1sin1 cot
= +
.
Nu thuc gc phn t I hoc II th 2
1sin1 cot
=+
.
Nu thuc gc phn t III hoc IV th 2
1sin1 cot
= +
.
II. Cho bit mt gi tr lng gic, tnh gi tr ca mt biu thc Cch 1: T GTLG bit, tnh cc GTLG c trong biu thc, ri thay vo biu thc. Cch 2: Bin i biu thc cn tnh theo GTLG bit III. Tnh gi tr mt biu thc lng gic khi bit tng hiu cc GTLG Ta thng s dng cc hng ng thc bin i: A B A B AB2 2 2( ) 2+ = + A B A B A B4 4 2 2 2 2 2( ) 2+ = +
A B A B A AB B3 3 2 2( )( )+ = + + A B A B A AB B3 3 2 2( )( ) = + + IV. Tnh gi tr ca biu thc bng cch gii phng trnh t t x t2sin , 0 1= x t2cos = . Th vo gi thit, tm c t. Biu din biu thc cn tnh theo t v thay gi tr ca t vo tnh. Thit lp phng trnh bc hai: t St P2 0 + = vi S x y P xy;= + = . T tm x, y. Bi 1. Cho bit mt GTLG, tnh cc GTLG cn li, vi:
a) a a0 04cos , 270 3605
= < < b) 2cos , 025
= < <
c) a a5sin ,13 2
= < < d) 0 01sin , 180 270
3 = < <
e) a a 3tan 3,2
= < < f) tan 2,2
= < <
g) 0cot15 2 3= + h) 3cot 3,2
= < <
Bi 2. Cho bit mt GTLG, tnh gi tr ca biu thc, vi:
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Trn S Tng Lng gic
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a) a aA khi a a
a acot tan 3sin , 0cot tan 5 2
+= =