题目内容
计算:
(1)
cos30°-
sin45° +tan60°
(2)
-2sin60° tan60° +2tan45°.
(1)
| 3 |
| 2 |
(2)
| sin30° |
| cos60° |
分析:(1)分别把cos30°=
,sin45°=
,tan60°=
代入原式进行计算即可;
(2)分别把sin30°=
,cos60°=
,sin60°=
,tan60°=
,tan45°=1代入进行计算.
| ||
| 2 |
| ||
| 2 |
| 3 |
(2)分别把sin30°=
| 1 |
| 2 |
| 1 |
| 2 |
| ||
| 2 |
| 3 |
解答:解:(1)原式=
×
-
×
+
=
-1+
=
+
;
(2)原式=
-2×
×
+2×1
=1-3+2
=0.
| 3 |
| ||
| 2 |
| 2 |
| ||
| 2 |
| 3 |
=
| 3 |
| 2 |
| 3 |
=
| 1 |
| 2 |
| 3 |
(2)原式=
| ||
|
| ||
| 2 |
| 3 |
=1-3+2
=0.
点评:本题考查的是特殊角的三角函数值,熟记各特殊角度的三角函数值是解答此题的关键.
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