当x∈[-6,-2/3]时,判断函数y=fx+f（x+2）的单调性

题目

当x∈[-6,-2/3]时,判断函数y=fx+f（x+2）的单调性
fx=2sin（π/4x+π/4） 我想看将fx带入后化简的步骤~

答案

f（x）=2sin（π/4x+π/4）
y=f(x)+f(x+2)
=2sin（π/4x+π/4）+2sin[π/4(x+2)+π/4]
=2sin（π/4x+π/4）+2sin[π/4x+π/2+π/4]
=2sin（π/4x+π/4）+2cos（π/4x+π/4)
=2√2[√√2/2*sin（π/4x+π/4）+√2/2*cos（π/4x+π/4)]
=2√2sin[(π/4*x+π/4)+π/4]
=2√2sin(π/4*x+π/2)
=2√2cos(π/4*x)
∵x∈[-6,-2/3]
∴-3π/2≤π/4x≤-π/6
∴当-3π/2≤π/4x≤-π,即x∈[-6,-4]时,函数递减,
当-π≤π/4x≤-π/6,即x∈[-4,-2/3]时,函数递增