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已知向量a=(2根号3sinx,cosx+sinx),b=(cosx,cosx-sinx),函数f(x)=a·b,求f(
(2)求f(x)的单调增区间,(3)若x属于【0,pai/2],求函数f(x)的值域
人气:196 ℃ 时间:2019-08-17 20:27:15
解答

a=(2√3sinx,cosx+sinx),b=(cosx,cosx-sinx)
f(x)=a·b=2√3sinxcosx+cos^2x-sin^2x
=√3sin2x+cos2x
=2sin(2x+π/6)
令2kπ-π/2求f(x)的解析式及周期f(x)=a·b=2√3sinxcosx+cos^2x-sin^2x=√3sin2x+cos2x=2sin(2x+π/6)周期T=2π/2=π
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