
已知函数f(x)=-√3sin^2x+sinxcosx 求在X∈[0,π/2]的值域
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f(x)=-√3sin^2x+sinxcosx
=-√3(1-cos2x)/2+sin2x/2
=sin2x/2+√3cos2x/2-√3/2
=sin(2x+π/3)-√3/2
因为X∈[0,π/2],所以2x+π/3∈[π/3,4π/3]
所以sin(2x+π/3)∈[-√3/2,1]
所以f(x)∈[-√3,1-√3/2]
=-√3(1-cos2x)/2+sin2x/2
=sin2x/2+√3cos2x/2-√3/2
=sin(2x+π/3)-√3/2
因为X∈[0,π/2],所以2x+π/3∈[π/3,4π/3]
所以sin(2x+π/3)∈[-√3/2,1]
所以f(x)∈[-√3,1-√3/2]
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