试题分析:(1) f(x)=a·b-1=(sin2x,2cosx)·(
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,cosx)-1
=
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sin2 x +2cos2 x -1=
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sin2x+cos2x=2sin(2x+
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) 4分
∴f(x)的最小正周期为π,最小值为-2. 6分
(2) f(
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)=2sin(
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+
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)=
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∴sin(
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+
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)=
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8分
∴
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+
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=
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∴ A=
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或
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(舍去) 10分
由余弦定理得a
2=b
2+c
2-2bccosA
52=64+c
2-8c即c
2-8c+12="0"
从而c=2或c=6 12分
点评:典型题,为研究三角函数的图象和性质,往往需要利用三角函数和差倍半公式将函数“化一”。本题由平面向量的坐标运算得到f(x)的表达式,通过“化一”,利用三角函数性质,求得周期、最小值。(2)则利用余弦定理,得到c的方程,达到解题目的。