Question

y=2sinxcosx_根号3cos2x+1的最小正周期和最值

最小值是?过程

Answer 1

y=2sinxcosx-√3cos2x+1
=2(1/2sin2x-√3/2cos2x)+1
=2(sin2x*cosπ/3-cos2xsinπ/3)+1
=2sin(2x+π/3)+1
T=π
当2x+π/3=π/2+2kπ,即x=π/12+kπ,k∈Z时，y取最大值为3
x=-5π/12+kπ,k∈Z时，y取最小值为-1

Answer 2

sin2x-√3cos2x+1=2(1/2sin2x-√3/2cos2x)+1=2sin(2x-π/3)+1
周期2π/2=π 最小值为-1，最大值为3