Question

函数y=1+(sinx+cosx)+(sinx+cosx)²的最大值

Answer 1

y=1+(sinx+cosx)+(sinx+cosx)²
= 1+√2sin(x+π/4)+[√2sin(x+π/4)]²
= 1+√2sin(x+π/4)+2sin²(x+π/4)
= 1+2[sin²(x+π/4)+√2/2sin(x+π/4)]
= 1+2[sin(x+π/4)+√2/4]²-1/4
= 2[sin(x+π/4)+√2/4]²+3/4
-1≤sin(x+π/4)≤1

-1+√2/4≤sin(x+π/4)+√2/4≤1+√2/4
0≤2[sin(x+π/4)+√2/4]²≤2(1+√2/4)²
3/4≤2[sin(x+π/4)+√2/4]²+3/4≤2(1+√2/4)²+3/4

最大值 = 2(1+√2/4)²+3/4 = 3+√2

Answer 2

令sinx+cosx=t，则y=t²+t+1(-√2＜t＜√2)
当t=√2时，ymax=3+√2，此时x=π/4+2kπ，k∈Z