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已知a=sin15°cos15°,b=cos2π6−sin2π6,c=tan30°1−tan230°,则a,b,c的大小关系是()A.a<b<cB.a>b>cC.c>a>bD.a<c<b

题目详情
已知a=sin15°cos15°,b=cos2
π
6
−sin2
π
6
c=
tan30°
1−tan230°
,则a,b,c的大小关系是(  )

A.a<b<c
B.a>b>c
C.c>a>b
D.a<c<b
b=cos2
π
6
−sin2
π
6
c=
tan30°
1−tan230°
,则a,b,c的大小关系是(  )

A.a<b<c
B.a>b>c
C.c>a>b
D.a<c<b
s2
π
6
−sin2
π
6
c=
tan30°
1−tan230°
,则a,b,c的大小关系是(  )

A.a<b<c
B.a>b>c
C.c>a>b
D.a<c<b
s2
π
6
−sin2
π
6
c=
tan30°
1−tan230°
,则a,b,c的大小关系是(  )

A.a<b<c
B.a>b>c
C.c>a>b
D.a<c<b
2
π
6
−sin2
π
6
c=
tan30°
1−tan230°
,则a,b,c的大小关系是(  )

A.a<b<c
B.a>b>c
C.c>a>b
D.a<c<b
π
6
ππ66n2
π
6
c=
tan30°
1−tan230°
,则a,b,c的大小关系是(  )

A.a<b<c
B.a>b>c
C.c>a>b
D.a<c<b
n2
π
6
c=
tan30°
1−tan230°
,则a,b,c的大小关系是(  )

A.a<b<c
B.a>b>c
C.c>a>b
D.a<c<b
2
π
6
c=
tan30°
1−tan230°
,则a,b,c的大小关系是(  )

A.a<b<c
B.a>b>c
C.c>a>b
D.a<c<b
π
6
ππ66c=
tan30°
1−tan230°
,则a,b,c的大小关系是(  )

A.a<b<c
B.a>b>c
C.c>a>b
D.a<c<b
tan30°
1−tan230°
tan30°tan30°1−tan230°1−tan230°tan230°tan230°230°




▼优质解答
答案和解析
∵a=sin15°cos15°=
1
2
sin30°=
1
4
b=cos2
π
6
−sin2
π
6
=cos
π
3
=
1
2
c=
tan30°
1−tan230°
=
1
2
tan60°=
3
2

∴a<b<c.
故选:A.
1
2
111222sin30°=
1
4
b=cos2
π
6
−sin2
π
6
=cos
π
3
=
1
2
c=
tan30°
1−tan230°
=
1
2
tan60°=
3
2

∴a<b<c.
故选:A.
1
4
111444,b=cos2
π
6
−sin2
π
6
=cos
π
3
=
1
2
c=
tan30°
1−tan230°
=
1
2
tan60°=
3
2

∴a<b<c.
故选:A.
b=cos2
π
6
−sin2
π
6
=cos
π
3
=
1
2
c=
tan30°
1−tan230°
=
1
2
tan60°=
3
2

∴a<b<c.
故选:A.
2
π
6
πππ666−sin2
π
6
=cos
π
3
=
1
2
c=
tan30°
1−tan230°
=
1
2
tan60°=
3
2

∴a<b<c.
故选:A.
2
π
6
πππ666=cos
π
3
=
1
2
c=
tan30°
1−tan230°
=
1
2
tan60°=
3
2

∴a<b<c.
故选:A.
π
3
πππ333=
1
2
c=
tan30°
1−tan230°
=
1
2
tan60°=
3
2

∴a<b<c.
故选:A.
1
2
111222,c=
tan30°
1−tan230°
=
1
2
tan60°=
3
2

∴a<b<c.
故选:A.
c=
tan30°
1−tan230°
tan30°tan30°tan30°1−tan230°1−tan230°1−tan230°230°=
1
2
tan60°=
3
2

∴a<b<c.
故选:A.
1
2
111222tan60°=
3
2

∴a<b<c.
故选:A.
3
2
3
3
3
3
33222,
∴a<b<c.
故选:A.