题目内容

已知α,β,γ成等差数列,且公差为
2π
3
,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______.
由题意,α=β-
2π
3
,γ=β+
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
[sin2(α+m)+sin2(β+m)+sin2(γ+m)]
=
1
3
[sin2(β-
2π
3
+m)+sin2(β+m)+sin2(β+
2π
3
+m)]
=
1
3
[
1-cos(2β-
4π
3
+2m)
2
+
1-cos(2β+2m)
2
+
1-cos(2β+
4π
3
+2m)
2
]
=
1
2
-
1
6
[cos(2β+2m-
4π
3
)+cos(2β+2m+
4π
3
)+cos(2β+2m)]
=
1
2
-
1
6
[2cos(2β+2m)cos
4π
3
+cos(2β+2m)]
=
1
2
-
1
6
[2cos(2β+2m)(-
1
2
)+cos(2β+2m)]
=
1
2
.
故答案为:
1
2
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