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求下列各式的值:(1)sin15°cos15°;(2)cos2π8-sin2π8;(3)tan22.5°1-tan222.5°;(4)2cos222.5°-1.

题目详情
求下列各式的值:
(1)sin15°cos15°;
(2)cos2
π
8
-sin2
π
8

(3)
tan22.5°
1-tan222.5°

(4)2cos222.5°-1.
求下列各式的值:
(1)sin15°cos15°;
(2)cos2
π
8
-sin2
π
8

(3)
tan22.5°
1-tan222.5°

(4)2cos222.5°-1.


2
π
8
-sin2
π
8

(3)
tan22.5°
1-tan222.5°

(4)2cos222.5°-1.
π
8
π8ππ882
π
8

(3)
tan22.5°
1-tan222.5°

(4)2cos222.5°-1.
π
8
π8ππ88
tan22.5°
1-tan222.5°

(4)2cos222.5°-1.
tan22.5°
1-tan222.5°
tan22.5°1-tan222.5°tan22.5°tan22.5°1-tan222.5°1-tan222.5°n222.5°n222.5°222.5°
2
▼优质解答
答案和解析
(1)sin15°cos15°=
1
2
sin30°=
1
4

(2)cos2
π
8
-sin2
π
8
=cos
π
4
=
2
2

(3)
tan22.5°
1-tan222.5°
=
1
2
tan45°=
1
2

(4)2cos222.5°-1=cos45°=
2
2
1
2
12111222sin30°=
1
4

(2)cos2
π
8
-sin2
π
8
=cos
π
4
=
2
2

(3)
tan22.5°
1-tan222.5°
=
1
2
tan45°=
1
2

(4)2cos222.5°-1=cos45°=
2
2
1
4
14111444;
(2)cos22
π
8
-sin2
π
8
=cos
π
4
=
2
2

(3)
tan22.5°
1-tan222.5°
=
1
2
tan45°=
1
2

(4)2cos222.5°-1=cos45°=
2
2
π
8
π8πππ888-sin22
π
8
=cos
π
4
=
2
2

(3)
tan22.5°
1-tan222.5°
=
1
2
tan45°=
1
2

(4)2cos222.5°-1=cos45°=
2
2
π
8
π8πππ888=cos
π
4
=
2
2

(3)
tan22.5°
1-tan222.5°
=
1
2
tan45°=
1
2

(4)2cos222.5°-1=cos45°=
2
2
π
4
π4πππ444=
2
2

(3)
tan22.5°
1-tan222.5°
=
1
2
tan45°=
1
2

(4)2cos222.5°-1=cos45°=
2
2
2
2
2
2
2
2
2
2
2
22222;
(3)
tan22.5°
1-tan222.5°
=
1
2
tan45°=
1
2

(4)2cos222.5°-1=cos45°=
2
2
tan22.5°
1-tan222.5°
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

(4)2cos222.5°-1=cos45°=
2
2
1
2
12111222tan45°=
1
2

(4)2cos222.5°-1=cos45°=
2
2
1
2
12111222;
(4)2cos2222.5°-1=cos45°=
2
2
2
2
2
2
2
2
2
2
2
22222.