题目内容
若f(n)=sin
,f(1)+f(3)+f(5)+…+f(101)=______.
| nπ |
| 6 |
因为y=sinx的周期是2π,
所以f(1)+f(3)+f(5)+…+f(11)
=sin
+sin
+sin
+sin
+sin
+sin
=
+1+
-
-1-
=0,
∴f(1)+f(3)+f(5)+…+f(101)
=8×(sin
+sin
+sin
+sin
+sin
+sin
)+sin
+sin
+sin
=sin
+sin
+sin
=
+1+
=2.
故答案为:2.
所以f(1)+f(3)+f(5)+…+f(11)
=sin
| π |
| 6 |
| 3π |
| 6 |
| 5π |
| 6 |
| 7π |
| 6 |
| 9π |
| 6 |
| 11π |
| 6 |
=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
∴f(1)+f(3)+f(5)+…+f(101)
=8×(sin
| π |
| 6 |
| 3π |
| 6 |
| 5π |
| 6 |
| 7π |
| 6 |
| 9π |
| 6 |
| 11π |
| 6 |
| π |
| 6 |
| 3π |
| 6 |
| 5π |
| 6 |
=sin
| π |
| 6 |
| 3π |
| 6 |
| 5π |
| 6 |
=
| 1 |
| 2 |
| 1 |
| 2 |
故答案为:2.
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