题目内容
计算:
(1)tan30°sin60°+cos230°-sin245°tan45°;
(2)
sin60°+2cos30°-
tan45°.
(1)tan30°sin60°+cos230°-sin245°tan45°;
(2)
| 1 |
| 2 |
| 3 |
分析:(1)求出每一部分的值,代入求出即可;
(2)求出每一部分的值,代入求出即可.
(2)求出每一部分的值,代入求出即可.
解答:解:(1)原式=
×
+(
)2-(
)2×1
=
+
-
=
;
(2)原式=
×
+2×
-
×1
=
+
-
=
.
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| 3 |
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| 2 |
| ||
| 2 |
| ||
| 2 |
=
| 1 |
| 2 |
| 3 |
| 4 |
| 1 |
| 2 |
=
| 3 |
| 4 |
(2)原式=
| 1 |
| 2 |
| ||
| 2 |
| ||
| 2 |
| 3 |
=
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| 4 |
| 3 |
| 3 |
=
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| 4 |
点评:本题考查了特殊角的三角函数值的应用,关键是能理解和记住各个特殊角的三角函数值.
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