题目内容
(文)已知函数f(x)=2sinx+3tanx.项数为27的等差数列{an}满足an∈(-
,
),且公差d≠0.若f(a1)+f(a2)+…+f(a27)=0,则当k值为
π |
2 |
π |
2 |
13
13
时有f(ak)=0.分析:利用和差化积和等差数列的性质可得sina13=0,进而可得f(a13)=0.
解答:解:∵f(a1)+f(a27)=2sina1+3tana1+2sina27+3tana27
=2(sina1+sina27)+3(
+
)
=4sin
cos
+3•
=4sina13cos13d+
=sina13(4cos13d+
),
同理f(a2)+f(a26)=sina13(4cos12d+
),
…,
f(a13)=2sina13+3tana13=sina13(2+
),
∵f(a1)+f(a2)+…+f(a27)=0,∴sina13=0.
∴f(a13)=0,
∴当k值为 13时有f(a13)=0.
故答案为0
=2(sina1+sina27)+3(
sina1 |
cosa1 |
sina27 |
cosa27 |
=4sin
a1+a27 |
2 |
a27-a1 |
2 |
sin(a1+a27) |
cosa1cosa27 |
=4sina13cos13d+
6sina13cosa13 |
cosa1cosa27 |
=sina13(4cos13d+
6cosa13 |
cosa1cosa27 |
同理f(a2)+f(a26)=sina13(4cos12d+
6cosa13 |
cosa2cosa26 |
…,
f(a13)=2sina13+3tana13=sina13(2+
3 |
cosa13 |
∵f(a1)+f(a2)+…+f(a27)=0,∴sina13=0.
∴f(a13)=0,
∴当k值为 13时有f(a13)=0.
故答案为0
点评:本题考查了和差化积和等差数列的性质,属于难题.
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