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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
1
2
的是(  )

A.sin15°cos15°
B.sin2
π
6
−cos2
π
6

C.2cos230°-1
D.
tan30°
1−tan230°
1
2
的是(  )

A.sin15°cos15°
B.sin2
π
6
−cos2
π
6

C.2cos230°-1
D.
tan30°
1−tan230°
1
2
1122


sin2
π
6
−cos2
π
6

C.2cos230°-1
D.
tan30°
1−tan230°
n2
π
6
−cos2
π
6

C.2cos230°-1
D.
tan30°
1−tan230°
n2
π
6
−cos2
π
6

C.2cos230°-1
D.
tan30°
1−tan230°
2
π
6
−cos2
π
6

C.2cos230°-1
D.
tan30°
1−tan230°
π
6
ππ66s2
π
6

C.2cos230°-1
D.
tan30°
1−tan230°
s2
π
6

C.2cos230°-1
D.
tan30°
1−tan230°
2
π
6

C.2cos230°-1
D.
tan30°
1−tan230°
π
6
ππ66
2
tan30°
1−tan230°
tan30°
1−tan230°
tan30°tan30°1−tan230°1−tan230°tan230°tan230°230°
▼优质解答
答案和解析
sin15°cos15°=
1
2
sin30°=
1
4
,故A错误;
sin2
π
6
-cos2
π
6
=-(cos2
π
6
-sin2
π
6
)=-cos
π
3
=-
1
2
,故B错误;
2cos230°-1=cos60°=
1
2
,故C正确;
tan30°
1−tan230°
=
1
2
2tan30°
1−tan230°
)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
1
2
111222sin30°=
1
4
,故A错误;
sin2
π
6
-cos2
π
6
=-(cos2
π
6
-sin2
π
6
)=-cos
π
3
=-
1
2
,故B错误;
2cos230°-1=cos60°=
1
2
,故C正确;
tan30°
1−tan230°
=
1
2
2tan30°
1−tan230°
)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
1
4
111444,故A错误;
sin22
π
6
-cos2
π
6
=-(cos2
π
6
-sin2
π
6
)=-cos
π
3
=-
1
2
,故B错误;
2cos230°-1=cos60°=
1
2
,故C正确;
tan30°
1−tan230°
=
1
2
2tan30°
1−tan230°
)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
π
6
πππ666-cos22
π
6
=-(cos2
π
6
-sin2
π
6
)=-cos
π
3
=-
1
2
,故B错误;
2cos230°-1=cos60°=
1
2
,故C正确;
tan30°
1−tan230°
=
1
2
2tan30°
1−tan230°
)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
π
6
πππ666=-(cos22
π
6
-sin2
π
6
)=-cos
π
3
=-
1
2
,故B错误;
2cos230°-1=cos60°=
1
2
,故C正确;
tan30°
1−tan230°
=
1
2
2tan30°
1−tan230°
)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
π
6
πππ666-sin22
π
6
)=-cos
π
3
=-
1
2
,故B错误;
2cos230°-1=cos60°=
1
2
,故C正确;
tan30°
1−tan230°
=
1
2
2tan30°
1−tan230°
)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
π
6
πππ666)=-cos
π
3
=-
1
2
,故B错误;
2cos230°-1=cos60°=
1
2
,故C正确;
tan30°
1−tan230°
=
1
2
2tan30°
1−tan230°
)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
π
3
πππ333=-
1
2
,故B错误;
2cos230°-1=cos60°=
1
2
,故C正确;
tan30°
1−tan230°
=
1
2
2tan30°
1−tan230°
)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
1
2
111222,故B错误;
2cos2230°-1=cos60°=
1
2
,故C正确;
tan30°
1−tan230°
=
1
2
2tan30°
1−tan230°
)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
1
2
111222,故C正确;
tan30°
1−tan230°
=
1
2
2tan30°
1−tan230°
)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
tan30°
1−tan230°
tan30°tan30°tan30°1−tan230°1−tan230°1−tan230°230°=
1
2
2tan30°
1−tan230°
)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
1
2
111222(
2tan30°
1−tan230°
)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
2tan30°
1−tan230°
2tan30°2tan30°2tan30°1−tan230°1−tan230°1−tan230°230°)=
1
2
tan60°=
3
2
,故D错误.
故选:C.
1
2
111222tan60°=
3
2
,故D错误.
故选:C.
3
2
3
3
3
3
33222,故D错误.
故选:C.