f﹙x－π／12﹚ = 2sin(3x - π/4+π/2) = 2sin(3x+π/4)
f(x-π/12)^2 = 4sin^2(3x+π/4) = 2(1-cos(6x+π/2)) = 2+2sin(6x)
x∈[﹣π／6,π／3],6x∈[﹣π,2π]
最大值4,6x = π／2,x = π／12我算的是f﹙x－π／12﹚＝2sin[3﹙x－π／12﹚＋π／2]＝2cos ﹙3x－π／4﹚＝2cos[﹙3x－3π／4﹚＋2π／4]＝﹣2sin﹙3x－3π／4﹚＝g﹙x﹚，请问我哪里算错了g=f的平方啊，所以用半角公式