题目内容
已知α,β,γ成等差数列,且公差为
,m为实常数,则sin2(α+m),sin2(β+m),sin2(γ+m)这三个三角函数式的算术平均数为______.
| 2π |
| 3 |
由题意,α=β-
,γ=β+
,
∴sin2(α+m),sin2(β+m),sin2(γ+m)这三个三角函数式的算术平均数为S=
[sin2(α+m)+sin2(β+m)+sin2(γ+m)]=
[sin2(β-
+m)+sin2(β+m)+sin2(β+
+m)]=
[
+
+
]=
-
[cos(2β+2m-
)+cos(2β+2m+
)+cos(2β+2m)]=
-
[2cos(2β+2m)cos
+cos(2β+2m)]=
-
[2cos(2β+2m)(-
)+cos(2β+2m)]=
.
故答案为:
| 2π |
| 3 |
| 2π |
| 3 |
∴sin2(α+m),sin2(β+m),sin2(γ+m)这三个三角函数式的算术平均数为S=
| 1 |
| 3 |
| 1 |
| 3 |
| 2π |
| 3 |
| 2π |
| 3 |
| 1 |
| 3 |
1-cos(2β-
| ||
| 2 |
| 1-cos(2β+2m) |
| 2 |
1-cos(2β+
| ||
| 2 |
| 1 |
| 2 |
| 1 |
| 6 |
| 4π |
| 3 |
| 4π |
| 3 |
| 1 |
| 2 |
| 1 |
| 6 |
| 4π |
| 3 |
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 2 |
| 1 |
| 2 |
故答案为:
| 1 |
| 2 |
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