题目内容
化简或求值:
①sin(x-y)siny-cos(x-y)cosy=
②sin70°cos10°-sin20°sin170°=
③cosα-
sinα=
④
=
⑤tan65°-tan5°=
⑥sin15°cos15°=
⑦sin2
-cos2
⑧2cos222.5°-1=
⑨
= .
①sin(x-y)siny-cos(x-y)cosy=
②sin70°cos10°-sin20°sin170°=
③cosα-
| 3 |
④
| 1+tan15° |
| 1-tan15° |
⑤tan65°-tan5°=
⑥sin15°cos15°=
⑦sin2
| θ |
| 2 |
| θ |
| 2 |
⑧2cos222.5°-1=
⑨
| 2tan150° |
| 1-tan2150° |
考点:三角函数中的恒等变换应用
专题:三角函数的求值
分析:直接结合三角恒等变换公式进行化简即可.
解答:
解:①sin(x-y)siny-cos(x-y)cosy
=-cos[(x-y)+y]=-cosx,
②sin70°cos10°-sin20°sin170°
=cos20°cos10°-siin20°sin10°
=cos(20°+10°)
=cos30°=
,
③cosα-
sinα=2(
cosα-
sinα)
=2cos(α+
),
④
=
=tan60°=
.
⑤tan65°-tan5°=tan(65°-5°)(1+tan65°tan5°)
=
(1+tan65°tan5°).
⑥sin15°cos15°=
sin30°=
,
⑦sin2
-cos2
=-cosθ
⑧2cos222.5°-1=cos45°=
,
⑨
=
=-
.
故答案为:①-cosx,②
;③2cos(α+
),④
;⑤
(1+tan65°tan5°);⑥
,⑦-cosθ;⑧
,⑨-
.
=-cos[(x-y)+y]=-cosx,
②sin70°cos10°-sin20°sin170°
=cos20°cos10°-siin20°sin10°
=cos(20°+10°)
=cos30°=
| ||
| 2 |
③cosα-
| 3 |
| 1 |
| 2 |
| ||
| 2 |
=2cos(α+
| π |
| 3 |
④
| 1+tan15° |
| 1-tan15° |
| tan45°+tan15° |
| 1-tan45°tan15° |
=tan60°=
| 3 |
⑤tan65°-tan5°=tan(65°-5°)(1+tan65°tan5°)
=
| 3 |
⑥sin15°cos15°=
| 1 |
| 2 |
| 1 |
| 4 |
⑦sin2
| θ |
| 2 |
| θ |
| 2 |
⑧2cos222.5°-1=cos45°=
| ||
| 2 |
⑨
| 2tan150° |
| 1-tan2150° |
2×(-
| ||||
1-(-
|
| 3 |
故答案为:①-cosx,②
| ||
| 2 |
| π |
| 3 |
| 3 |
| 3 |
| 1 |
| 4 |
| ||
| 2 |
| 3 |
点评:本题综合考查了三角恒等变换公式、二倍角公式等知识,属于中档题.
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