设函数f(x)=2sinxcos^2φ/2+cosxsinφ-sinx(0

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设函数f(x)=2sinxcos^2φ/2+cosxsinφ-sinx(0

设函数f(x)=2sinxcos^2φ/2+cosxsinφ-sinx(0
设函数f(x)=2sinxcos^2φ/2+cosxsinφ-sinx(0<φ<π)在x=π处取得最小值.求f(x)的单调递增区间

设函数f(x)=2sinxcos^2φ/2+cosxsinφ-sinx(0
f(x)=2sinxcos^2(φ/2)+cosxsinφ-sinx
=sinx*[2cos^2(φ/2) -1] +cosxsinφ
=sinxcosφ +cosxsinφ
=sin(x+φ)
由于f(x)在x=π处有最小值,则sin(π+φ)=-1
即sinφ=1
因为0

设函数f(x)=2sinxcos^2(φ/2)+cosxsinφ-sinx(0<φ<π)在x=π处取得最小值.求f(x)的单调递增区间
f(x)=sinx(1+cosφ)+cosxsinφ-sinx=sinxcosφ+cosxsinφ=sin(x+φ)
f(π)=sin(π+φ)=-sinφ=-1,故φ=π/2,于是f(x)=sin(x+π/2)=-cosx
故其单调递减区间为[2kπ-π,2kπ],k∈Z.

f(x)=sinx(cosφ+1)+cosxsinφ-sinx
=sinxcosφ+cosxsinφ
=sin(x+φ)
因为在x=π处取得最小值,所以φ=π/2,
所以2kπ-π/2 <=x+π/2<=2kπ+π/2,
单调递增区间为[2kπ-π,2kπ]