已知函数f(x)=(1+1/tanx)sin2x-2sin(x+π/4)sin(x-π/4)1.若tanx=2,求f(a) 2.若x∈[π/12,π/2],求f(x)的取值范围
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![已知函数f(x)=(1+1/tanx)sin2x-2sin(x+π/4)sin(x-π/4)1.若tanx=2,求f(a) 2.若x∈[π/12,π/2],求f(x)的取值范围](/uploads/image/z/5349912-24-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%281%2B1%2Ftanx%29sin2x-2sin%28x%2B%CF%80%2F4%29sin%28x-%CF%80%2F4%291.%E8%8B%A5tanx%3D2%2C%E6%B1%82f%EF%BC%88a%EF%BC%89+2.%E8%8B%A5x%E2%88%88%5B%CF%80%2F12%2C%CF%80%2F2%5D%2C%E6%B1%82f%28x%29%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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已知函数f(x)=(1+1/tanx)sin2x-2sin(x+π/4)sin(x-π/4)1.若tanx=2,求f(a) 2.若x∈[π/12,π/2],求f(x)的取值范围
已知函数f(x)=(1+1/tanx)sin2x-2sin(x+π/4)sin(x-π/4)
1.若tanx=2,求f(a) 2.若x∈[π/12,π/2],求f(x)的取值范围
已知函数f(x)=(1+1/tanx)sin2x-2sin(x+π/4)sin(x-π/4)1.若tanx=2,求f(a) 2.若x∈[π/12,π/2],求f(x)的取值范围
将函数f(X)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π/4) 化简得:=(1+cosx/sinx)*2sinxcosx+m(sinxcosπ/4+cosxsinπ/4)(sinxcosπ/4-cosxsinπ/4) =2sinxcosx+2(cosx)^2+1/2[(sinx)^2-(cosx)^2]m 由tana=2得 sina=2cosa 所以:f(a)=(sina)^2+2(cosa)^2+1/2*3m(cosa)^2 =1+(cosa)^2+3m/2*(cosa)^2 =3/5 所以得:3m/2*(cosa)^2+2/5[(cosa)^2+(sina)^2]+(cosa)^2=0 3m/2*(cosa)^2+7/5*(cosa)^2+2/5*(sina)^2=0 因为tana=2 将上式同除于(cosa)^2得((cosa)^2由tana=2知不会为零):3m/2+7/5+2/5*4=0 所以:m=-2