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Ưng dụng của lượng giác.. Dạng 1:Bài toán tam giác.[r]
(1)Ưng dụng lượng giác.
Dạng 1:Bài toán tam giác
VD1:cmr:mọi tam giác ta lncó: 1/SinA+sinB+sinC=4cos 2.cos 2.cos
C B A
Giai:
2sin cos 2sin cos 2sin (cos cos B A B
A B A B
A B A B
A B
A
=4sin .cos 2 cos 2 4sin 2.sin 2.sin C B A B
A A B A B A B A B
2/cosA²+cosB²+cosC²=1-2cosA.cosB.cosC Giai:
VT: cos( )
2 cos
2 cos
B A B
A
²=1+ cos( )
2 cos
cos
B A B
A
=1+cos(A+B)(cos(A B)cos(AB) =1-2.cosA.cosB.cosC=vp
3/ sin2A+sin2B+sin2C=4sinA.sinB.sinC
Vi:sin2C=sin2(ð-(A+B))=sin(2ð-2(A+B))=-sin2(A+B)
VT:2sin(A+B)(cos(A-B)-cos(A+B))=2sinC.(-2)sinA.sin(-B)=4sinAsinBsinC.=vp
3/(truong hop tam giac khong vuong) tanA+tanB+tanC=tanA.tanB.tanC
Ta co:tan(A+B)=tan(ð-C)=-tanC=
B A
B A
tan tan
tan tan
dpcm 4/tan 2tan tan tan tan 2tan 1
C A C
B B
A
Giai:
taco:tan( 2)tan( B
A
ð/2-
B A
B A
C C
tan tan
2 tan tan
2 tan
1 )
(2)5/
) cot cot cot tan tan (tan sin
1 sin
1 sin
1 A B C A B C
C B
A
ta di CM:cot cot cot cot 2.cot 2.cot C B A C
B A
Bang cach tuong tu nhu tren )
2 cot cot cot tan tan (tan
: A B C A B C
VP
= 2(tan cot 2)
1 ) cot (tan ) cot (tan
1 A A B B C C
=
vp C B
A C
C B
B A
A sin
1 sin
1 sin
1 cos sin
2
2 cos sin
2 cos sin
2
2 cos cos
2 sin
2 cos cos
2 sin
2 cos cos
2 sin
B A
C
C A
B
C B
A
6/
ta co: cos2.cos2.cos2
2 cos sin ) cos cos cos
sin sin
( 2 cos cos
2 sin
2 cos cos
2 sin
C B A
B A B A
C B A
B A
C A
B
C B
A
=
2 tan tan
cos cos
2 sin sin cos
cos A B
B A
B A B
A
De dang cm duoc
2 tan tan cos cos
2
sin A B
B A
C
dpcm
BAI TAP VAN DUNG:
(3)sin
2 tan tan tan tan tan tan sin sin sin cos cos sin cos cos sin cos cos
C A C
B B
A C
B A B
A C C
A B C
B A
2/ cho tam giac ABC cm:
2 ^
2 ^ ^ sin
) sin(
c b a C
B
A
3/ Trong tam giac ABC biet tan tan B A
= 2( )
1 :
b a c
CM
4/ CM moi tam giac ta co:
.cotA +cotB+cotC= S
c b a
4
2 ^ ^
^
5/ cm rang moi tam giac bat ki ta co :
tan a a
a
a a
h r
h h
r h C B
2 2
tan
6/ cho tam giac ABC cm:
bc.cot )
3 1 ( ^ cot cot
2 Rp a b c p
C ab B ac A
7/ cmr:voi moi tam giac ABC ta luon co:
1+R A B C
r
cos cos
cos
8/ CMR voi moi tam giac ta luon co: a.cotA+b.cotB+c.cotC=2(r+R)
9/ cmr voi moi tam giac ABC ta luon co:
cosAcosBcosC≤8
10/ cmr voi tam giac ABC bat ki:1<cosA+cosB+cosC≤3/2
(4)Dạng 1:Bài toán tam giác
VD1:cmr:mọi tam giác ta lncó: 1/SinA+sinB+sinC=4cos 2.cos 2.cos
C B A
Giai:
2sin cos 2sin cos 2sin (cos cos B A B
A B A B
A B A B
A B
A
=4sin .cos 2 cos 2 4sin 2.sin 2.sin C B A B
A A B A B A B A B
2/cosA²+cosB²+cosC²=1-2cosA.cosB.cosC Giai:
VT: cos( )
2 cos
2 cos
B A B
A
²=1+ cos( )
2 cos
cos
B A B
A
=1+cos(A+B)(cos(A B)cos(AB) =1-2.cosA.cosB.cosC=vp
3/ sin2A+sin2B+sin2C=4sinA.sinB.sinC
Vi:sin2C=sin2(ð-(A+B))=sin(2ð-2(A+B))=-sin2(A+B)
VT:2sin(A+B)(cos(A-B)-cos(A+B))=2sinC.(-2)sinA.sin(-B)=4sinAsinBsinC.=vp
3/(truong hop tam giac khong vuong) tanA+tanB+tanC=tanA.tanB.tanC
Ta co:tan(A+B)=tan(ð-C)=-tanC=
B A
B A
tan tan
tan tan
dpcm 4/tan 2tan tan tan tan 2tan 1
C A C
B B
A
Giai:
taco:tan( 2)tan( B
A
ð/2-
B A
B A
C C
tan tan
2 tan tan
2 tan
1 )
(5)) cot cot cot tan tan (tan sin
1 sin
1 sin
1 A B C A B C
C B
A
ta di CM:cot cot cot cot 2.cot 2.cot C B A C
B A
Bang cach tuong tu nhu tren )
2 cot cot cot tan tan (tan
: A B C A B C
VP
= 2(tan cot 2)
1 ) cot (tan ) cot (tan
1 A A B B C C
=
vp C B
A C
C B
B A
A sin
1 sin
1 sin
1 cos sin
2
2 cos sin
2 cos sin
2
2 cos cos
2 sin
2 cos cos
2 sin
2 cos cos
2 sin
B A
C
C A
B
C B
A
6/
ta co: cos2.cos2.cos2
2 cos sin ) cos cos cos
sin sin
( 2 cos cos
2 sin
2 cos cos
2 sin
C B A
B A B A
C B A
B A
C A
B
C B
A
=
2 tan tan
cos cos
2 sin sin cos
cos A B
B A
B A B
A
De dang cm duoc
2 tan tan cos cos
2
sin A B
B A
C
dpcm
BAI TAP VAN DUNG:
(6)sin
2 tan tan tan tan tan tan sin sin sin cos cos sin cos cos sin cos cos
C A C
B B
A C
B A B
A C C
A B C
B A
2/ cho tam giac ABC cm:
2 ^
2 ^ ^ sin
) sin(
c b a C
B
A
3/ Trong tam giac ABC biet tan tan B A
= 2( )
1 :
b a c
CM
4/ CM moi tam giac ta co:
.cotA +cotB+cotC= S
c b a
4
2 ^ ^
^
5/ cm rang moi tam giac bat ki ta co :
tan a a
a
a a
h r
h h
r h C B
2 2
tan
6/ cho tam giac ABC cm:
bc.cot )
3 1 ( ^ cot cot
2 Rp a b c p
C ab B ac A
7/ cmr:voi moi tam giac ABC ta luon co:
1+R A B C
r
cos cos
cos
8/ CMR voi moi tam giac ta luon co: a.cotA+b.cotB+c.cotC=2(r+R)
9/ cmr voi moi tam giac ABC ta luon co:
cosAcosBcosC≤8
10/ cmr voi tam giac ABC bat ki:1<cosA+cosB+cosC≤3/2