(1)已知函数f(x)=cos^2x+2sinxcosx-sin^2x.若f(α/2)=3/4,试求sin2α的值(2)若不等式-sin2x+tcos2x≥0在区间(π/12,π/6]上恒成立,求实数t的取值
(1)已知函数f(x)=cos^2x+2sinxcosx-sin^2x.若f(α/2)=3/4,试求sin2α的值(2)若不等式-sin2x+tcos2x≥0在区间(π/12,π/6]上恒成立,求实数t的取值
数学人气:918 ℃时间:2020-03-23 22:02:58
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f(x)=cos^2x+2sinxcosx-sin^2x=cos^2x-sin^2x+2sinxcosx=cos2x+sin2xf(α/2)=cosα+sinα=3/4(cosα+sinα)²=(3/4)²1+sin2α=9/16sin2α=9/16-1=-7/16 sin2x+tcos2x=√(1+t²)sin(2x+φ)因为x∈...谢谢,那第二题呢第二题:sin2x+tcos2x=√(1+t²)sin(2x+φ)因为x∈(π/12,π/6] 所以2x∈(π/6,π/3] 要使sin2x+tcos2x≥0在区间(π/12,π/6]上恒成立必须使sin(2x+φ)≥0在区间(π/12,π/6]上恒成立即需要π/6+φ≥0且π/3+φ≤π即-π/6≤φ≤2π/3又cosφ=1/√(1+t²),且cosφ在[-π/6,2π/3]内的值域是[√3/2,1]所以有√3/2≤1/√(1+t²)≤1解得0≤t≤√3/3或者 -√3/3≤t≤0
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