已知函数f(x)=(1+1/tanx)sin(x)^2 -2sin(x+π/4)*cos(x+π/4)
(1)若tanα=2,求f(α)
(2)若x∈[π/12,π/2],求f(x)的取值范围
人气:191 ℃ 时间:2019-08-20 14:40:02
解答
先用tanx=sinx/cosx、倍角公式、诱导公式化简原函数:
f(x)=sin²x+sinxcosx-sin[2(x+π/4)]=(1-cos2x)/2+1/2sin2x-sin(2x+π/2)=-3/2cos2x+1/2sin2x+1/2
(1)由万能公式,cos2x=(1-tan²x)/(1+tan²x),sin2x=2tanx/(1+tan²x)
f(α)=-3/2*(1-2²)/(1+2²)+1/2*2*2/(1+2²)=13/10
(2)为求f(x)范围,再用辅助角公式化简原函数:
f(x)=根号10/2sin(2x-arctan3)+1/2
若x∈[π/12,π/2]:f(x)最大值为(根号10)/2+1/2
f(x)最小值为f(π/12)=-3/2cos(π/6)+1/2sin(π/6)+1/2=-(3根号3)/4+3/4
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