题目内容
计算:
(1)6cos30°×tan30°-2sin245°;
(2)3
-(π-1)0-2sin45°+tan45°.
(1)6cos30°×tan30°-2sin245°;
(2)3
| 2 |
(1)原式=6×
×
-2×(
)2,
=3-2×
,
=2.
(2)3
-(π-1)0-2sin45°+tan45°,
=3
-1-2×
+1,
=3
-1-
+1,
=2
.
| ||
| 2 |
| ||
| 3 |
| ||
| 2 |
=3-2×
| 1 |
| 2 |
=2.
(2)3
| 2 |
=3
| 2 |
| ||
| 2 |
=3
| 2 |
| 2 |
=2
| 2 |
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