Question

求y=(a+sinx)(a+cosx)(a属于一切实数）的最小值 高一数学

Answer 1

y=(a+sinx)(a+cosx)
=sinxcosx+a(sinx+cosx)+a^2
=(1/2)sin2x+√2sin(x+π/4)+a^2
=-(1/2)cos(2x+π/2)+√2sin(x+π/4)+a^2
=-(1/2){1-2[sin(x+π/4)]^2}+√2sin(x+π/4)+a^2
=sin(x+π/4)]^2+√2sin(x+π/4)+a^2-1/2
令t=sin(x+π/4) 则有-1<=t<=1
y=t^2+√2t+a^2-1/2
由-1<=t<=1知
当且仅当t=-√2/2时，y有最小值a^2-1
此时，x=2k+5π/4或2k+7π/4（k∈Z）

Answer 2

y= a²+acosx+asinx+sinxcosx=a²+sinxcosx+a（sinx+cosx）
设sinx+cosx=t t∈[-√2,√2] 即y=a²+（t²-1）/2+t
y=t²/2+t+a²-1/2=0.5(t+1)²+a²-1 t=-√2时 ymin=a²-√2
不懂追问