题目内容
计算:
(1)
+(
) -
-π0+2log36-log312-2ln
(2)sin120°•cos330°-sin210°•cos(-60°)+tan675°.
(1)
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| 64 |
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| e |
(2)sin120°•cos330°-sin210°•cos(-60°)+tan675°.
分析:(1)通过根式、分式以及对数的运算法则化简求值即可.
(2)直接利用诱导公式以及两角和与差的三角函数化简求值即可.
(2)直接利用诱导公式以及两角和与差的三角函数化简求值即可.
解答:解:(1)
+(
) -
-π0+2log36-log312-2ln
=
+
-1+log336-log312-2ln
=
+
-1+1-1=2
(2)原式=sin(180°-60°)•cos(360°-30°)-sin(180°+30°)cos(60°)+tan(675°-720°)=sin60°cos30°+sin30°cos60°+tan(-45°)
=
×
+
×
-tan45°
=
+
-1=0.
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| e |
=
| 5 |
| 3 |
| 4 |
| 3 |
| e |
=
| 5 |
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| 3 |
(2)原式=sin(180°-60°)•cos(360°-30°)-sin(180°+30°)cos(60°)+tan(675°-720°)=sin60°cos30°+sin30°cos60°+tan(-45°)
=
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| 2 |
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| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
=
| 3 |
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| 1 |
| 4 |
点评:本题(1)考查根式与分数指数幂的互化及其化简运算以及对数运算法则,基本知识的考查.(2)考查特殊角的三角函数,运用诱导公式化简求值,考查计算能力.
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