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下列各式中,值为12的是()A.sin15°•cos15°B.2cos2π12-1C.1+cos30°2D.tan22.5°1−tan222.5°

题目详情
下列各式中,值为
1
2
的是(  )

A.sin15°•cos15°
B.2cos2
π
12
-1
C.
1+cos30°
2

D.
tan22.5°
1−tan222.5°
1
2
的是(  )

A.sin15°•cos15°
B.2cos2
π
12
-1
C.
1+cos30°
2

D.
tan22.5°
1−tan222.5°
1
2
1122


2
π
12
-1
C.
1+cos30°
2

D.
tan22.5°
1−tan222.5°
π
12
ππ1212
1+cos30°
2

D.
tan22.5°
1−tan222.5°
1+cos30°
2
1+cos30°
2
1+cos30°
2
1+cos30°
2
1+cos30°1+cos30°22
tan22.5°
1−tan222.5°
tan22.5°
1−tan222.5°
tan22.5°tan22.5°1−tan222.5°1−tan222.5°tan222.5°tan222.5°222.5°
▼优质解答
答案和解析
A,∵sin15°•cos15°=
1
2
sin30°=
1
4
1
2

B,∵2cos2
π
12
-1=cos
π
6
=
3
2
1
2

C,∵
1+cos30°
2
=
2+
3
4
1
2

D,∵
tan22.5°
1−tan222.5°
=
1
2
tan45°=
1
2

故选:D.
1
2
111222sin30°=
1
4
1
2

B,∵2cos2
π
12
-1=cos
π
6
=
3
2
1
2

C,∵
1+cos30°
2
=
2+
3
4
1
2

D,∵
tan22.5°
1−tan222.5°
=
1
2
tan45°=
1
2

故选:D.
1
4
111444≠
1
2

B,∵2cos2
π
12
-1=cos
π
6
=
3
2
1
2

C,∵
1+cos30°
2
=
2+
3
4
1
2

D,∵
tan22.5°
1−tan222.5°
=
1
2
tan45°=
1
2

故选:D.
1
2
111222;
B,∵2cos22
π
12
-1=cos
π
6
=
3
2
1
2

C,∵
1+cos30°
2
=
2+
3
4
1
2

D,∵
tan22.5°
1−tan222.5°
=
1
2
tan45°=
1
2

故选:D.
π
12
πππ121212-1=cos
π
6
=
3
2
1
2

C,∵
1+cos30°
2
=
2+
3
4
1
2

D,∵
tan22.5°
1−tan222.5°
=
1
2
tan45°=
1
2

故选:D.
π
6
πππ666=
3
2
1
2

C,∵
1+cos30°
2
=
2+
3
4
1
2

D,∵
tan22.5°
1−tan222.5°
=
1
2
tan45°=
1
2

故选:D.
3
2
3
3
3
3
33222≠
1
2

C,∵
1+cos30°
2
=
2+
3
4
1
2

D,∵
tan22.5°
1−tan222.5°
=
1
2
tan45°=
1
2

故选:D.
1
2
111222;
C,∵
1+cos30°
2
=
2+
3
4
1
2

D,∵
tan22.5°
1−tan222.5°
=
1
2
tan45°=
1
2

故选:D.
1+cos30°
2
1+cos30°
2
1+cos30°
2
1+cos30°
2
1+cos30°1+cos30°1+cos30°222=
2+
3
4
1
2

D,∵
tan22.5°
1−tan222.5°
=
1
2
tan45°=
1
2

故选:D.
2+
3
4
2+
3
4
2+
3
4
2+
3
4
2+
3
2+
3
2+
3
3
33444≠
1
2

D,∵
tan22.5°
1−tan222.5°
=
1
2
tan45°=
1
2

故选:D.
1
2
111222;
D,∵
tan22.5°
1−tan222.5°
=
1
2
tan45°=
1
2

故选:D.
tan22.5°
1−tan222.5°
tan22.5°tan22.5°tan22.5°1−tan222.5°1−tan222.5°1−tan222.5°222.5°=
1
2
tan45°=
1
2

故选:D.
1
2
111222tan45°=
1
2

故选:D.
1
2
111222.
故选:D.