Question

已知0<a≤√2,求函数y=(sinx+a)(cosx+a)的最值

Answer 1

设sinx+cosx=t，则t=sinx+cosx=√2[(√2/2)sinx+(√2/2)cosx]=√2sin(x+π/4)
∴t∈[-√2，√2]
∵t²=(sinx+cosx)²=sin²x+cos²x+2sinxcosx=1+2sinxcosx
∴sinxcosx=(t²-1)/2则
y=(sinx+a)(cosx+a)=sinxcosx+a(sinx+cosx)+a²=(t²-1)/2+at+a²
=(1/2)(t+a)²+(a²-1)/2
对称轴为t= -a
∵-√2≤t≤√2，-√2≤ -a＜0
∴当t= -a时，y取最小值(a²-1)/2
当t=√2时，y取最大值a²+√2a+ (1/2)

Answer 2

y=(sinx+a)(cosx+a) = sinx*cosx + a(sinx+cosx) +a^2
设 t =sinx +cosx, 则 t^2 = sin^2x +cos^2x + 2sinx*cosx
sinx*cosx = ( t^2 -1)/2, 而且 t = √2*sin(x+pi/4). 所以 - √2 <=t <=√2
y= (t^2 -1)/2 +t +a^2 = (1/2)* [ t^2 + 2t +1] -1 + a^2= (1/2)*(t+1)^2 +a^2 -1
当 t =-1 时，ymin = a^2,-1
当 t=√2时，ymax = a^2 +1/2 +√2