Question

求函数f(x)=sin^2x+sinxcosx的最值与最小正周期

Answer 1

解：
f(x)=(1-cos2x)/2+(sin2x)/2
=(1/2)sin2x-(1/2)cos2x+1/2
=(√2/2)*[sin2x*cos(π/4)-cos(2x)sin(π/4)]+1/2
=(√2/2)sin(2x-π/4)+1/2
T=2π/2=π，
最大值为(√2/2)+(1/2)
最小值为-(√2/2)+(1/2)

Answer 2

f(x)=sin^2x+sinxcosx
=(1/2)(1-cos2x)+(1/2)sin2x
=(1/2)(sin2x-cos2x)+1/2
=(√2/2)sin(2x-π/4)+1/2
最大值为 √2/2+1/2
最小值为 -√2/2+1/2

最小正周期为 2π/2=π

Answer 3

f(x)=sin^2x+sinxcosx
=(1/2)(1-cos2x)+(1/2)sin2x
=(1/2)(sin2x-cos2x)+1/2
=(√2/2)sin(2x-π/4)+1/2
最大值为 √2/2+1/2
最小值为 -√2/2+1/2

最小正周期为 2π/2=π