题目内容
计算:(
)
+9log3
-sin
| 1 |
| 32 |
| 1 |
| 5 |
| 2 |
| 7π |
| 6 |
3
3
.分析:(
)
=
,9log3
=2,sin
=-sinπ,由此能求出(
)
+9log3
-sin
的值.
| 1 |
| 32 |
| 1 |
| 5 |
| 1 |
| 2 |
| 2 |
| 7π |
| 6 |
| 1 |
| 32 |
| 1 |
| 5 |
| 2 |
| 7π |
| 6 |
解答:解:(
)
+9log3
-sin
=
+2+sin
=
+2+
=3.
故答案为:3.
| 1 |
| 32 |
| 1 |
| 5 |
| 2 |
| 7π |
| 6 |
=
| 1 |
| 2 |
| π |
| 6 |
=
| 1 |
| 2 |
| 1 |
| 2 |
=3.
故答案为:3.
点评:本题考查指数的运算,是基础题.解题时要认真审题,注意正弦函数的应用.
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