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下列各式中,值为12的是()A.sin15°cos15°B.sin2π6−cos2π6C.2cos230°-1D.tan30°1−tan230°
题目详情
下列各式中,值为
的是( )
A.sin15°cos15°
B.sin2
−cos2
C.2cos230°-1
D.
的是( )
A.sin15°cos15°
B.sin2
−cos2
C.2cos230°-1
D.
1 1 2 2
sin2
−cos2
C.2cos230°-1
D.
n2
−cos2
C.2cos230°-1
D.
n2
−cos2
C.2cos230°-1
D.
2
−cos2
C.2cos230°-1
D.
π π 6 6 s2
C.2cos230°-1
D.
s2
C.2cos230°-1
D.
2
C.2cos230°-1
D.
π π 6 6
2
tan30° tan30° 1−tan230° 1−tan230° tan230°tan230°230°
1 |
2 |
A.sin15°cos15°
B.sin2
π |
6 |
π |
6 |
C.2cos230°-1
D.
tan30° |
1−tan230° |
1 |
2 |
A.sin15°cos15°
B.sin2
π |
6 |
π |
6 |
C.2cos230°-1
D.
tan30° |
1−tan230° |
1 |
2 |
sin2
π |
6 |
π |
6 |
C.2cos230°-1
D.
tan30° |
1−tan230° |
π |
6 |
π |
6 |
C.2cos230°-1
D.
tan30° |
1−tan230° |
π |
6 |
π |
6 |
C.2cos230°-1
D.
tan30° |
1−tan230° |
π |
6 |
π |
6 |
C.2cos230°-1
D.
tan30° |
1−tan230° |
π |
6 |
π |
6 |
C.2cos230°-1
D.
tan30° |
1−tan230° |
π |
6 |
C.2cos230°-1
D.
tan30° |
1−tan230° |
π |
6 |
C.2cos230°-1
D.
tan30° |
1−tan230° |
π |
6 |
2
tan30° |
1−tan230° |
tan30° |
1−tan230° |
▼优质解答
答案和解析
sin15°cos15°=
sin30°=
,故A错误;
sin2
-cos2
=-(cos2
-sin2
)=-cos
=-
,故B错误;
2cos230°-1=cos60°=
,故C正确;
=
(
)=
tan60°=
,故D错误.
故选:C.
1 1 12 2 2sin30°=
,故A错误;
sin2
-cos2
=-(cos2
-sin2
)=-cos
=-
,故B错误;
2cos230°-1=cos60°=
,故C正确;
=
(
)=
tan60°=
,故D错误.
故选:C.
1 1 14 4 4,故A错误;
sin22
-cos2
=-(cos2
-sin2
)=-cos
=-
,故B错误;
2cos230°-1=cos60°=
,故C正确;
=
(
)=
tan60°=
,故D错误.
故选:C.
π π π6 6 6-cos22
=-(cos2
-sin2
)=-cos
=-
,故B错误;
2cos230°-1=cos60°=
,故C正确;
=
(
)=
tan60°=
,故D错误.
故选:C.
π π π6 6 6=-(cos22
-sin2
)=-cos
=-
,故B错误;
2cos230°-1=cos60°=
,故C正确;
=
(
)=
tan60°=
,故D错误.
故选:C.
π π π6 6 6-sin22
)=-cos
=-
,故B错误;
2cos230°-1=cos60°=
,故C正确;
=
(
)=
tan60°=
,故D错误.
故选:C.
π π π6 6 6)=-cos
=-
,故B错误;
2cos230°-1=cos60°=
,故C正确;
=
(
)=
tan60°=
,故D错误.
故选:C.
π π π3 3 3=-
,故B错误;
2cos230°-1=cos60°=
,故C正确;
=
(
)=
tan60°=
,故D错误.
故选:C.
1 1 12 2 2,故B错误;
2cos2230°-1=cos60°=
,故C正确;
=
(
)=
tan60°=
,故D错误.
故选:C.
1 1 12 2 2,故C正确;
=
(
)=
tan60°=
,故D错误.
故选:C.
tan30° tan30° tan30°1−tan230° 1−tan230° 1−tan230°230°=
(
)=
tan60°=
,故D错误.
故选:C.
1 1 12 2 2(
)=
tan60°=
,故D错误.
故选:C.
2tan30° 2tan30° 2tan30°1−tan230° 1−tan230° 1−tan230°230°)=
tan60°=
,故D错误.
故选:C.
1 1 12 2 2tan60°=
,故D错误.
故选:C.
3 3 32 2 2,故D错误.
故选:C.
1 |
2 |
1 |
4 |
sin2
π |
6 |
π |
6 |
π |
6 |
π |
6 |
π |
3 |
1 |
2 |
2cos230°-1=cos60°=
1 |
2 |
tan30° |
1−tan230° |
1 |
2 |
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
1 |
2 |
1 |
4 |
sin2
π |
6 |
π |
6 |
π |
6 |
π |
6 |
π |
3 |
1 |
2 |
2cos230°-1=cos60°=
1 |
2 |
tan30° |
1−tan230° |
1 |
2 |
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
1 |
4 |
sin22
π |
6 |
π |
6 |
π |
6 |
π |
6 |
π |
3 |
1 |
2 |
2cos230°-1=cos60°=
1 |
2 |
tan30° |
1−tan230° |
1 |
2 |
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
π |
6 |
π |
6 |
π |
6 |
π |
6 |
π |
3 |
1 |
2 |
2cos230°-1=cos60°=
1 |
2 |
tan30° |
1−tan230° |
1 |
2 |
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
π |
6 |
π |
6 |
π |
6 |
π |
3 |
1 |
2 |
2cos230°-1=cos60°=
1 |
2 |
tan30° |
1−tan230° |
1 |
2 |
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
π |
6 |
π |
6 |
π |
3 |
1 |
2 |
2cos230°-1=cos60°=
1 |
2 |
tan30° |
1−tan230° |
1 |
2 |
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
π |
6 |
π |
3 |
1 |
2 |
2cos230°-1=cos60°=
1 |
2 |
tan30° |
1−tan230° |
1 |
2 |
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
π |
3 |
1 |
2 |
2cos230°-1=cos60°=
1 |
2 |
tan30° |
1−tan230° |
1 |
2 |
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
1 |
2 |
2cos2230°-1=cos60°=
1 |
2 |
tan30° |
1−tan230° |
1 |
2 |
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
1 |
2 |
tan30° |
1−tan230° |
1 |
2 |
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
tan30° |
1−tan230° |
1 |
2 |
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
1 |
2 |
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
2tan30° |
1−tan230° |
1 |
2 |
| ||
2 |
故选:C.
1 |
2 |
| ||
2 |
故选:C.
| ||
2 |
3 |
3 |
3 |
故选:C.
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