Question

已知函数y=sinx+cosx+2sinxcosx,求函数y的最大值

已知函数y=sinx+cosx+2sinxcosx,求函数y的最大值。

Answer 1

sinx+cosx=√2(sinxcosπ/4+cosxsinπ/4)=√2sin(x+π/4)

y = sinx+cosx+2sinxcosx
= sinx+cosx+2sinxcosxsin^2x+cos^2x-1
= sin^2x+2sinxcosx+cos^2x + sinx+cosx - 1
= (sinx+cosx)^2+(sinx+cosx)-1
= [√2sin(x+π/4)]^2 + √2sin(x+π/4) - 1
= 2[sin(x+π/4)]^2 + √2sin(x+π/4) - 1
= 2[sin(x+π/4)+√2/4]^2 - 5/4
-1≤sin(x+π/4)≤1
(-4+√2)/4 ≤ sin(x+π/4)+√2/4 ≤ (4+√2)/4
0 ≤ [sin(x+π/4)+√2/4)^2 ≤ (9+4√2)/8
0 ≤2 [sin(x+π/4)+√2/4)^2 ≤ (9+4√2)/4
-5/4 ≤2 [sin(x+π/4)+√2/4)^2 - 5/4 ≤ 1+√2
y的最大值 1+√2