Khóa luận tốt nghiệp toán học: Một số bài toán nhận dạng tam giác - Pdf 14

class="bi x0 y0 w1 h1"
∆ABC ABC
A, B, C A, B, C
a, b, c
A, B, C
h
a
, h
b
, h
c
A, B, C
m
a
, m
b
, m
c
A, B, C
l
a
, l
b
, l
c
A, B, C
R
r
r
a
, r

2R
a
2
= b
2
+ c
2
− 2bc cos A = (b −c)
2
+ 4bc sin
2
A
2
b
2
= c
2
+ a
2
− 2ca cos B = (c − a)
2
+ 4ca sin
2
B
2
c
2
= a
2
+ b

tan
C−A
2
tan
C+A
2
a
2
= b
2
+ c
2
− 4S cot A; b
2
= c
2
+ a
2
− 4S cot B; c
2
= a
2
+ b
2
− 4S cot C
cot A =
b
2
+ c
2

cot
A
2
+ cot
C
2

c = a cos B + b cos A = r

cot
A
2
+ cot
B
2

m
2
a
=
b
2
+ c
2
2

a
2
4
; m

;
h
b
= c sin A = a sin C =
2S
b
;
h
c
= a sin B = b sin A =
2S
c
l
a
=
2bc
b + c
cos
A
2
=
2

bc
b + c

p(p − a)
l
b
=
































(p − a)r
a
= (p − b)r
b
= (p − c)r
c

p(p − a)(p − b)(p −c)
R =











a
2 sin A
=
b
2 sin B
=
c
2 sin C
abc
4S

A
2
=
b sin
C
2
sin
A
2
cos
B
2
=
c sin
A
2
sin
B
2
cos
C
2
4R sin
A
2
. sin
B
2
. sin
C

2
cos
B
2
= p. tan
B
2
r
c
=
S
p − c
=
c cos
A
2
cos
B
2
cos
C
2
= p. tan
C
2
ABC
sin(A + B) = sin C; cos(A + B) = −cos C
sin
A + B
2

2
sin A + sin B + sin C = 2 sin
A + B
2
. cos
A − B
2
+ 2 sin
C
2
. cos
C
2
= 2 cos
C
2

cos
A − B
2
+ cos
A + B
2

= 4 cos
A
2
. cos
B
2

cos
A − B
2
− cos
A + B
2

= 1 + 4 sin
A
2
. sin
B
2
. sin
C
2
cos 2A + cos 2B + cos 2C = −1 − 4 cos A. cos B. cos C
cos 2A + cos 2B + cos 2C = 2 cos(A + B). cos(A −B) + 2 cos
2
C − 1
= −1 − 2 cos C[cos(A − B) + cos(A + B)]
= −1 − 4 cos A. cos B. cos C
tan A + tan B + tan C = tan A. tan B. tan C
tan(A + B) = −tan C
tan A + tan B
1 − tan A. tan B
= −tan C.
tan A + tan B + tan C = tan A. tan B. tan C
tan
A

tan
A
2
+ tan
B
2
1 − tan
A
2
tan
B
2
=
1
tan
C
2
,
tan
A
2
tan
B
2
+ tan
B
2
tan
C
2

2
= cot
A
2
. cot
B
2
. cot
C
2
tan
A
2
. tan
B
2
+ tan
B
2
. tan
C
2
+ tan
C
2
. tan
A
2
= 1,
1

C
2
= cot
A
2
cot
B
2
cot
C
2
sin
2
A + sin
2
B + sin
2
C = 2 + 2 cos A. cos B. cos C
sin
2
A + sin
2
B + sin
2
C =
3
2

1
2

; c
2
= ac

1
h
2
=
1
b
2
+
1
c
2
a
2
= b
2
+ c
2
A
B C
H
h
a
b
c
c


n
n (a
1
, a
2
, , a
n
) (b
1
, b
2
, , b
n
)(n ≥ 2)
(a
1
b
1
+ a
2
b
2
+ + a
n
b
n
)
2
≤ (a
2

n
(a
n
) (b
n
).
a
1
+ a
2
+ + a
n
n
.
b
1
+ b
2
+ + b
n
n

a
1
b
1
+ a
2
b
2

.
b
1
+ b
2
+ + b
n
n

a
1
b
1
+ a
2
b
2
+ + a
n
b
n
n
ABC A, B, C a, b, c
BC, CA, AB
|a − b| < c < a + b; |b − c| < a < b + c; |c − a| < b < c + a
a ≥ b
A ≥ B
∆ABC
a) cos 2A +


5
2
= 0

2 cos A −

3 cos(B − C)

2
+ 3 sin
2
(B −C) = 0,



sin(B − C) = 0
cos A =

3
2
cos(B − C) = 0

A = 30
0
B = C = 75
0
ABC A = 30
0
; B = C = 75
0

+
1
2
+ 2 cos
2
A

,
2(1 + cos A)
2
≤ 3(1 + 2 cos
2
A).

1 + 2 cos
2
A ≥
2

2

3
cos
2
A
2
,

1 + 2 cos
2

2
B)(1 + 2 cos
2
C), T ≥ 0
T ≥

2

2

3

3
cos
2
A
2
cos
2
B
2
cos
2
C
2
.

1 + 2 cos
2
A

+

1 + 2 cos
2
C
sin A


6
3

cot
A
2
cot
B
2
cot
C
2
,
cot
A
2
cot
B
2
cot
C
2

> 0
cot
A
2
cot
B
2
cot
C
2
≥ 3

3.

1 + 2 cos
2
A
sin B
+

1 + 2 cos
2
B
sin C
+

1 + 2 cos
2
C
sin A

ABC A = B = C = 60
0
ABC
cos 2A + 2

2 cos B + 2

2 cos C = 3
ABC.
cos
2
A + 2

2 sin
A
2
. cos
B − C
2
− 2 = 0,
cos A(cos A −1) +

1 − 2 sin
2
A
2

+ 2

2 sin

B − C
2

2
−sin
2
B − C
2
≤ 0,









cos A = 0

2 sin
A
2
= cos
B − C
2
sin
B − C
2
= 0

3
2
,
cos
C − A
2
=

3
2
= cos
π
6
.
C > A ABC









C − A
2
=
π
6
C + A =

ABC
sin
2
A + sin
2
B =
2n+1

sin
2
C
A B C.
C
C > 90
0
cos C ∈ (−1; 0) cos C =
a
2
+ b
2
− c
2
2ab
a
2
+ b
2
< c
2
.

B <
2n+1

sin
2
C.
C < 90
0
2n+1

sin
2
C < 1.
sin
2
A + sin
2
B = 1 + cos C. cos(A −B).
C < 90
0
A, B cos C > 0 cos(A − B) > 0.
2n+1

sin
2
C = sin
2
A + sin
2
B = 1 + cos C. cos(A − B) > 1

π
3

= 0,
2 sin

C
2

π
6

−cos
A − B
2
+ cos

C
2

π
6

= 0,




sin


3
B =
π
3
C =
π
3
ABC 60
0
.
ABC
sin A + sin B + sin C =
r + 4R sin
2
C
2
2R sin
C
2
C = 120
0
ABC r = 4R sin
A
2
sin
B
2
sin
C
2

,
4 cos
A
2
. cos
B
2
. cos
C
2
= 2

sin
A
2
sin
B
2
+ cos
A
2
cos
B
2
− sin
A
2
sin
B
2

C − 1
= −2 cos A. cos B. cos C
M = 0 cos A. cos B. cos C = 0. ∆ABC
M < 0 cos A. cos B. co s C > 0. cos A, cos B, cos C
∆ABC
cos A > 0, cos B > 0 cos C > 0. ABC
M > 0 cos A. cos B. cos C < 0.
ABC
ABC
a)

b
2
+ c
2
≤ a
2
sin A + sin B + sin C = 1 +

2
b)

cos A + cos B + cos C =

2
cos
2
A + cos
2
B + cos

.
sin A + sin B + sin C = sin A + 2 cos
A
2
. cos
B − C
2
≤ 1 + 2.


2
2

.1 = 1 +

2,
sin A + sin B + sin C = 1 +

2







sin A = 1
cos
A
2

cos
2
A+cos
2
B+cos
2
C ≥ 1 cos A. cos B. co s C ≤ 0.
ABC A ≥ 90
0
B, C < 90
0
.
1 −

2 = 2 sin
2
A
2
− 2 sin
A
2
. cos
B − C
2
≥ 2 sin
2
A
2
− 2 sin
A


2
2
f(x) = x
2
−x f

(x) = 2x−1 > 0 x ∈


2
2
; 1

,
f(x) x ∈


2
2
; 1

.
f


2
2

=

= 1

A = 90
0
B = C = 45
0
ABC A = 90
0
; B = C = 45
0
ABC
a) p tan
B
2
tan
C
2
= p − c
b)
r
R
= 2 sin
2
C
2
+
1
4
cos
2


1 − cos
2
C
2
=

(p − a)(p − b)
ab
.
tan
C
2
=

(p − a)(p − b)
p(p − c)
,
tan
B
2
=

(p − a)(p − c)
p(p − b)
.
p

(p − a)(p − b)
p(p − c)

= cos A + co s B + cos C,
r
R
= −2 sin
2
C
2
+ 2 sin
C
2
. cos
B − C
2
.
4 sin
2
C
2
− 2 sin
C
2
. cos
B − C
2
+
1
4
= 0,
1
4


1
2
cos
B − C
2
= 0



B = C
sin
C
2
=
1
4
ABC C.
ABC
a) h
a
=

bc. cos
A
2
b) m
a
=




bc.
h
a


bc. cos
A
2
,
b = c.
ABC A.
m
2
a
=
1
4
(b
2
+ c
2
+ 2bc cos A).
1
4
(b
2
+ c
2

sin A − sin B + sin C
sin A + sin B + sin C
= tan
B
2
. tan
C
2
ABC
4 cos
2
C
2
2 sin
C
2
. cos
C
2
=
sin(A + B)
cos A. cos B
2 cos A. cos B = 2 sin
2
C
2
cos(A+B)+cos(A−B) = 1−cos C, cos(A−B) = 1,
A = B.
ABC C.
2 sin

A
2
. sin
C
2
cos
A
2
. cos
C
2
= tan
B
2
. tan
C
2
,
tan
A
2
= tan
B
2
A = B.
ABC C.
ABC A, B, C <
π
2
.

A + B
2
,
(tan A + tan B)
2
≥ 4 tan
2
A + B
2
,
(tan A − tan B)
2
≤ 0.
tan A = tan B A = B
ABC C.
ABC
a) 2 cos B. sin A. sin C
+

3

sin B + 2 sin
2
B
2
+ 4 sin
A
2
. sin
B

2 sin A. sin C
,
2 cos B. sin A. sin C = sin
2
A + sin
2
C − sin
2
B,
cos A + cos B + cos C − 1 = 4 sin
A
2
. sin
B
2
. sin
C
2
,
2 sin
2
B
2
= 1 − cos B.
sin
2
A + sin
2
C − sin
2

2

2
= 0,















cos A =

3
2
cos C =

3
2
sin B =

3

cos
B + C
2
. cos
B − C
2
=
19
9
,
4 sin
2
A
2

4
3
sin
A
2
. cos
B − C
2
+
1
9
= 0,

2 sin
A

1
3
cos
B − C
2
= 0



B = C
sin
A
2
=
1
6
ABC A.
ABC
a)
1 + cos B
sin B
=
2 sin A + sin C

4 sin
2
A − sin
2
C
b)

sin B −sin C = sin A(cos B −cos C),
sin
B − C
2

cos
B + C
2
+ sin A. sin
B + C
2

= 0.
cos
B + C
2
+ sin A. sin
B + C
2
> 0
sin
B − C
2
= 0,
B = C.
ABC A.
ABC
a) sin
A
2


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