题目内容
已知向量
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(1)求函数g(x)的最小正周期;
(2)在△ABC中,a,b,c分别是角A,B,C的对边,且f(c)=3,c=1,

【答案】分析:(1)根据向量的数量积表示出函数g(x)的解析式,然后根据余弦函数的二倍角公式降幂化为y=Acos(wx+ρ)的形式,根据T=
可得答案.
(2)先根据向量的数量积表示出函数f(x)的解析式,然后化简为y=Asin(wx+ρ)的形式,将C代入函数f(x),根据f(c)=3求出C的值,再由余弦定理可求出a,b的值.
解答:解:(Ⅰ)g(x)=
=1+sin22x=1+
=-
cos4x+
∴函数g(x)的最小周期T=
(Ⅱ)f(x)=
=2
=cos2x+1+
sin2x=2sin(2x+
)+1
f(C)=2sin(2C+
)+1=3∴sin(2C+
)=1
∵C是三角形内角∴2C+
,∴2C+
即:C=
∴cosC=
=
即:a2+b2=7
将ab=2
可得:
解之得:a2=3或4
∴a=
或2∴b=2或
,∵a>b,∴a=2 b=
点评:本题主要考查三角函数最小正周期的求法和余弦定理的应用.属基础题.
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(2)先根据向量的数量积表示出函数f(x)的解析式,然后化简为y=Asin(wx+ρ)的形式,将C代入函数f(x),根据f(c)=3求出C的值,再由余弦定理可求出a,b的值.
解答:解:(Ⅰ)g(x)=

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∴函数g(x)的最小周期T=

(Ⅱ)f(x)=


=cos2x+1+
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f(C)=2sin(2C+
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∵C是三角形内角∴2C+
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∴cosC=
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将ab=2
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∴a=
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点评:本题主要考查三角函数最小正周期的求法和余弦定理的应用.属基础题.

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