Question

已知f(x)=sin(2x+π/6)+sin(2x-π/6)+2[(cos)^2]x+a，当x属于【-π/4，π/4】时，f（x）的最小值为-3，求a

Answer 1

f(x)=sin(2x+π/6)+sin(2x-π/6)+2[(cos)^2]x+a
=2sin2xcos(π/6)+cos2x+1 +a
=2[sin2xcos(π/6)+cos2xsin(π/6)]+1 +a
=2sin(2x+π/6)+1+a
令 -π/2+2kπ≤2x+π/6≤π/2+2kπ
解得 -π/3+kπ≤x≤π/6+kπ
由于 x∈[-π/4，π/4]，所以f(x)在[-π/4，π/6]上是增函数，
同理，f(x)在[π/6，π/4]上是减函数。
所以 最大值为f(π/6)，最小值为f(-π/4)=-3
即 2sin(-π/2 +π/6)+1+a=-3，a=√3-4
解：由和差化积得：f(x)=sin(2x+π/6)+sin(2x-π/6)+2[(cos)^2]x+a=2sin2xcos(π/6)+cos2x+1 +a
=2[sin2xcos(π/6)+cos2xsin(π/6)]+1 +a
=2sin(2x+π/6)+1+a
x属于【-π/4，π/4】时 => -π/2《2x《π/2 => -π/3《2x+π/6《2/3π
由此可以知道f(x) 在-π/3《2x+π/6《2/3π区间内单调递增，所以当x=-π/3 时，函数有最小值-3
得：-3=2sin[2*(-π/3)+π/6]+1+a
-3=2sin[-2/3π+π/6]+1+a
-3=2sin[-π/2]+1+a
-3=2*(-1)+1+a
-3=-2+1+a
a=-3+2-1
a=-2

Answer 2

f(x)=sin(2x+π/6)+sin(2x-π/6)+2[(cos)^2]x+a
=2sin2xcos(π/6)+cos2x+1 +a
=2[sin2xcos(π/6)+cos2xsin(π/6)]+1 +a
=2sin(2x+π/6)+1+a
令 -π/2+2kπ≤2x+π/6≤π/2+2kπ
解得 -π/3+kπ≤x≤π/6+kπ
由于 x∈[-π/4，π/4]，所以f(x)在[-π/4，π/6]上是增函数，
同理，f(x)在[π/6，π/4]上是减函数。
所以 最大值为f(π/6)，最小值为f(-π/4)=-3
即 2sin(-π/2 +π/6)+1+a=-3，a=√3-4

Answer 3

解：由和差化积得：f(x)=sin(2x+π/6)+sin(2x-π/6)+2[(cos)^2]x+a=2sin2xcos(π/6)+cos2x+1 +a
=2[sin2xcos(π/6)+cos2xsin(π/6)]+1 +a
=2sin(2x+π/6)+1+a
x属于【-π/4，π/4】时 => -π/2《2x《π/2 => -π/3《2x+π/6《2/3π
由此可以知道f(x) 在-π/3《2x+π/6《2/3π区间内单调递增，所以当x=-π/3 时，函数有最小值-3
得：-3=2sin[2*(-π/3)+π/6]+1+a
-3=2sin[-2/3π+π/6]+1+a
-3=2sin[-π/2]+1+a
-3=2*(-1)+1+a
-3=-2+1+a
a=-3+2-1
a=-2