f(x)=3sinωxcosωx+√3cos2ωx-√3/2 +1 求函数y=f(x)值域f(x)=3sinωxcosωx+√3cos^2ωx-√3/2 +1 求函数y=f(x)值域
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f(x)=3sinωxcosωx+√3cos2ωx-√3/2 +1 求函数y=f(x)值域f(x)=3sinωxcosωx+√3cos^2ωx-√3/2 +1 求函数y=f(x)值域
f(x)=3sinωxcosωx+√3cos2ωx-√3/2 +1 求函数y=f(x)值域
f(x)=3sinωxcosωx+√3cos^2ωx-√3/2 +1 求函数y=f(x)值域
f(x)=3sinωxcosωx+√3cos2ωx-√3/2 +1 求函数y=f(x)值域f(x)=3sinωxcosωx+√3cos^2ωx-√3/2 +1 求函数y=f(x)值域
f(x)=3sinωxcosωx+√3cos^2ωx-√3/2 +1
=3/2sin2ωx+√3/2cos2ωx+1
=√3sin(2ωx+π/6)-√3/2+1
因此,值域为:[-3√3/2+1,√3/2+1]
f(x)=3sinωxcosωx+(√3)cos2ωx-√3/2 +1 求函数y=f(x)值域
f(x)=(3/2)sin2ωx+(√3)cos2ωx-√3/2 +1
=(3/2)[sin2ωx+(2√3/3)cos2ωx]-√3/2 +1
=(3/2)[sin2ωx+tanφcos2ωx]-√3/2 +1
=[3/(...
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f(x)=3sinωxcosωx+(√3)cos2ωx-√3/2 +1 求函数y=f(x)值域
f(x)=(3/2)sin2ωx+(√3)cos2ωx-√3/2 +1
=(3/2)[sin2ωx+(2√3/3)cos2ωx]-√3/2 +1
=(3/2)[sin2ωx+tanφcos2ωx]-√3/2 +1
=[3/(2cosφ)][sin2ωxcosφ+cos2ωxsinφ]-√3/2 +1
=[(3/2)√21]sin(2ωx+φ)-√3/2 +1
其中tanφ=(2/3)√3,φ∊(0,π/2);sinφ=2√(1/7),cosφ=1/√21.
故值域为[-(3/2)√21-√3/2 +1 ,(3/2)√21-√3/2 +1 ]
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