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已知f(x)= √3sinx+cosx (1)求f(x)最小正周期及最大值 (2)若f(α)=2/3,求cos(2α+π/3)的值
人气:148 ℃ 时间:2020-06-19 19:25:02
解答
f(x)= √3sinx+cosx
=2[(√3/2)sinx+(1/2)cosx]
=2sin(x+π/6)
(1) 最小正周期T=2π/1=2π
∵sin(x+π/6)≤1
∴f(x)最大值=2
(2) f(α)=2sin(α+π/6)=2/3
sin(α+π/6)=1/3
cos(2α+π/3)=cos[2(α+π/6)]
=1-2[sin(α+π/6)]^2
=1-2*(1/3)^2
=1-2/9
=7/9
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