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题型一 :三角恒等变形与三角函数图象问题
b2025·cd·efYghij
0
w >
,
2
2
2 3
s
c
in
os
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x
x
w
w
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.
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y
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=
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(2)s
1
w = t,u
0,2π
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y
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si (
n
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y
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3
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a
c
B
b
C
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c =
,
2
a
b
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V
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(3)k
2
b
a
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sin 2A
B
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π
2sin π
sin 2
f x
x
x
æ
ö
=
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π
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6
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x
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题型二 :三角形中边长及周长问题
1¬b2025·ÈÉÊË·Nehij , ,
a b c ª¹o
ABC
V
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A B C '·¸,º
3 cos
sin
3
0
a
C
c
A
b
+
-
=
.
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(2)k
3, sin
2 3
c
a
B
=
=
,p a .
2¬b24-25 (·ÌÍÍη±)9hA
ABC
V
5,
, ,
A B C Ï·'¸ª¹o , ,
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ij
2
3
cos
3 cos
a
c
B
b
C
-
=
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(1)p B 'Xr
(2)k
3
c =
,
2
a
b
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ABC
V
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2
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ABC
V
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2¬b2025·íî·efYghï
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V
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V
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,
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1
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2
3
BD
CD =
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题型三 :三角形中面积问题
1¬b2025·Ìõ·Nehuö÷
3sin ,sin
cos
m
x
x
x
=
+
r
,
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cos
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x
x
x
=
-
r
,
f x
m n
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(1)p
f x '¿6ÀÁÂÃr
(2)AÐ
ABC
V
5, A
B
C
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b
c
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1
f
A = ,
2
a =
,
6
sin
sin
2
B
C
+
=
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ABC
V
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ABC
V
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A
C
=
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2
2
2
mac
a
c
b
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-
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m 'qr
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3
b =
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ABC
V
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ABC
V
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sin
)(
)
(sin
sin
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A
B b
a
c
B
C
+
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ABC
V
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ABC
V
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3sin
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C
C
b =
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2
2
2
2
b
c
ac
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.
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bⅱhk
5 2
a
c
+
=
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ABC
V
'»¼S .
题型四 :解三角形中三线问题
1¬b2025·ąĆÑć·ehA
ABC
V
5, , ,
a b c ª¹o , ,
A B C Ï·'¸,º
1
cos
2 c
b
a
C
=
-
, A 'ĈªÅ¦ BC Ç D,º
2
BD
DC
=
¬
(1)p Ar
(2)k
3
AC =
,p AD 'Ѭ
2¬b24-25 (·ÈÉ·±)9hA
ABC
V
5, A , B ,C ,Ï·¸ª¹o a ,b ,
c ,ij cos
sin
cos
sin
a
A
a
A
b
B
b
B
+
=
+
,º
.
a
b
¹
(1)p
;
C
(2)k D o AB ¸'5/,º
1
AB = ,
5
2
CD =
,p
ABC
V
'»¼.
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NÜ-áG»¼b»¼ªĊh,ÔÕ-ĈªÅKL M.
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(ÅR,NWáGHIKÔÕ-»¼L M.
1¬b2025·ċ°ČηNehï
ABC
V
'¶ A,B,C '·¸o a,b,c,ij
2
2
2
2
b
c
a
bc
+
-
=
,
2sin
sin .
C
A
B
-
=
(1)psin ;
C
(2)u
10
BC =
,p BC ¸c'(.
2¬b24-25 (·ă°ÑĄ·²³´µhA
ABC
V
5, , ,
A B C Ï·'¸ª¹o , ,
a b c ,čĎ
3
(sin
c
b
A
=
+
3cos )
A .
(1)p B 'Xr
(2)k
ABC
V
'»¼o
3
2
,
B
Ð
'ĈªÅ BD ¦ AC Ç/ D ,º BD = 1,p c
a 'q.
3¬b24-25 (c·ă½ďù·nĐhA
ABC
V
5, A,B,C Ï·'¸ª¹o a,b,c,/ D oų AC '
5/,A,C čĎ
2
2
(sin
sin )
sin
sin sin .
A
C
A
C
A
C
-
=
+
-
(1)p Br
(2)k
ABC
V
'»¼o 3 ,
13
b =
,p5ŠBD 'Ѭ
题型五:三角形中图形类边长及范围问题
1¬b24-25 (·ċ°·²³´µhAÐ! ABC 5, A B C ·đ'¸ª¹
o a b c ,ij 2 sin
sin
3 sin
a
B
C
b
A
=
.
(1)pC r
(2)p b
a 'zq{|.
2¬b2025·øõĒĆ·ehA
ABC
V
5, A , B ,C Ï·'¸ª¹o a ,b ,c .ij
2sin cos
sin
2sin
C
A
A
B
=
+
.
(1)pC 'Xr
(2)p cos
cos
B
A 'zq{|.
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1¬b2025·ĘÎęĚ·efYghij
ABC
V
'¶ A B C '·¸ª¹o a ,b ,c ,
º 3 cos
cos
cos
b
B
a
C
c
A
=
+
¬
(1)p tanBr
(2)k
π π
,
4 3
A
é
ù
Îê
ú
ë
û ,º
1
a = ,pb
c
+ 'zq{|¬
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ABC
V
5, A,B,C '·¸ª¹o a,b,c,
2
A
B
=
.
(1)ðñ
2
a
b b
c
=
+
r
(2)k
ABC
V
-Ð!,p
c
a
b
-
'zq{|.
题型六:三角形中证明类问题
b24-25 (c·ĝĞ·nĐhu
*
n
n
n
A B C
nÎN
ğ
'¶
,
,
n
n
n
A B C '·¸ª¹o
,
,
n
n
n
a b c ,
ij
1
1
1
1
1
,
2
b
c b
c
a
>
+
=
¬
(1)p
1A 'zq{|r
(2)k·Ġì'
*
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2
na =
,º
1
,
,
n
n
n
a b
c
+
æĢ
ģ,
1
,
,
n
n
n
a c
b
+
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ģ,ðñ
n
n
n
A B C
V
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ÑoKq¬
b24-25 (·ĝĞ·²³´µhA
ABC
V
5, , ,
A B C '·¸ª¹o
, , ,2
2 cos
0
a b c
a
b
C
c
-
+
=
¬
(1)p B r
(2)k
3,
7,
a
b
D
=
=
o BC ¸cN/,º
ABD
△
'»¼o 5 3
4
,ðñ
1
3
BD
BC
=
¬
题型七:解三角形中内切圆、外接圆问题
1¬b24-25 (·ċ½Ĥη²³´µhA
ABC
V
5, ABC '·¸ª¹o ab
c,º
2
2
2
1
3
a
b
c
-
=
.
(1)k
ABC
V
'»¼
2
1
6
S
c
=
,p Ar
(2)k tan
3
C =
,
ABC
V
'»¼
6
S = ,p
ABC
V
'ýĥĦ'»¼.
2¬b24-25 (·ċ½ħĨĩ·±)9hij
ABC
V
5, , ,
A B C '·¸ª¹-
, ,
a b c , 3
sin
3 cos
a
c
B
b
C
-
=
.
(1)ðñ
, ,
A B C æĢ
ģr
(2)k
7
b =
,
ABC
V
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1
s
2
ABC
C
r
D
=
×
·ÇýĥĭīĬR NWáG
HIK
2R
sin
sin
sin
=
=
=
C
c
B
b
A
a
M.
1¬b24-25 (·Èõ·±)9hij
ABC
V
'¶ A, B ,C Ï·'¸ª¹o a ,
b ,c ,
2
2
sin
cos2
sin
1
A
B
C
+
+
= ,º a ,b ,c æ
ģ¬
(1)p B r
(2)k/ D čĎ AB
BD
=
uuur
uuur ,
ABC
V
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3
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BCD
△
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2¬b24-25 (·ċ½ıK·±)9hA
ABC
V
5,ij , ,
A B C '·¸ª¹o , , ,
a b c
BAC
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BC Ç/ D ,
,
,
ABD
ACD
ABC
V
V
V
'ýĥĦ'īĬª¹o
1
2
R R
R
,
,
,º
1
2
R
R
R
+
=
.
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1
2 sin 2
BC
R
BAC
Ð
=
r
(2)p
BAC
Ð
r
(3)k
2 3
a =
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题型八:解三角形中图形类问题
1¬b24-25 (·IJijĄĴĵ·±)9hij
ABC
V
'¶ A , B ,C '·¸ª¹-
a ,b ,c ,ij
2
2
2
2
2
4
cos
ac
c
B
a
b
c
+
=
-
-
¬
(1)p B ;
(2)k D o
ABC
V
ýN/,A͏! ABCD 5,¸Ñ
2,
,
30
BC
DCB
B
CAD
=
Ð
= Ð
Ð
=
o , p¸CD 'lXq¬
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ABC
V
5,Q o¸ BC cN/,
2
BQ
CQ
=
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,
AB AC c'ķ/bäēĸĹ/h,k
120
MQN
Ð
=
o ,ºu
.
CNQ
q
Ð
=
(1)pð
äĺq oĻq, QM
QN oKq.
(2)s BMQ
V
v CNQ
V
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AC
^
,
AB
BC
^
, AC Ĉª
BCD
Ð
.
(1)k
π
6
ACD
Ð
=
,
2
CD =
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(2)k BD
CD
=
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BCD
Ð
.
2¬b24-25 (c·ÈÉŁł·nĐhÞ,Ĉ»Ķ¸! ABCD 5
AD
AC
AB
BC
^
^
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,
,AC Ĉª
.
BCD
Ð
(1)k
30
2
ACD
CD
Ð
=
=
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CD
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,
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bⅱhpsin
.
BDC
Ð
N B
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ABC
V
5,
3
AB = ,
7
BC =
,
3 21
sin
14
C =
(1)p
A
Ð
r
(2)k D o AC '5/,pcos
DBC
Ð
.
2¬b24-25 (c·ĘÎŅɰ·nĐhA
ABC
V
5, , ,
A B C '·¸ª¹o , ,
a b c ,º
3 sin
cos
2
a
B
b
A
b
+
=
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(2)k
2
a =
,p
ABC
V
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ABC
V
5,ij
2 3,
3
3,
3 2
AB
BC
AC
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=
+
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.
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V
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BE
+
'zq{|.
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3cos2
1
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x
x
=
-
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f x '¿6ÀÁÂÃr
(2)k
6
π
π
5
,
2
x
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2
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m
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m 'zq{|¬
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ABC
V
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2 cos
a
c
b
C
-
=
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3
b
A =
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ABC
Ð
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2
BD = ,p
ABC
V
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π
3
A =
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BAC
Ð
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AB
BD
AC
DC
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r
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2
c =
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1
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a b c ª¹o
ABC
V
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A B C '·¸,º 3 sin
cos
a
C
a
C
b
c
+
-
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o BC ¸'5/,k
3
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2
2
2
4
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a
-
=
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(2)p
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V
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2
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π
3
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V
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6
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3
a
B
b
A
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10
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A
B
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.
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a b c ,º
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B
b
A
c
A
+
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4,
6
b
c
=
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ABC
V
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CA
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q
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π
3
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