一道简单高一数学题

y = 3tan(π/6-x/4) = -3tan(x/4-π/6)
最小正周期T = π/ω = π/(1/4) = 4π
因为y=tanx单调增区间为x∈[kπ-π/2,kπ+π/2](k∈Z)
又因为A<0所以y=3tan(π/6-x/4)单调减区间为
kπ-π/2≤x/4-π/6≤kπ+π/2](k∈Z)
即4kπ-4/3π≤x≤4kπ+8/3π](k∈Z)
所以y=3tan(π/6-x/4)单调减区间为x∈[4kπ-4/3π,4kπ+8/3π]](k∈Z)
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详细说明:
y=3tan(π/6-x/4)中A=3 ω=-1/4 φ=π/6
因为ω<0所以用诱导公式把y=3tan(π/6-x/4)化成y=-3tan(x/4-π/6)
最小正周期T = π/ω 而对于y=A sin或者cos(ωx + φ)则是T = 2π/ω
之后因为A<0所以y=tanx单调增区间为x∈[kπ-π/2,kπ+π/2]](k∈Z)对应的kπ-π/2≤x/4-π/6≤kπ+π/2](k∈Z)是y=-3tan(x/4-π/6)是减区间,A>0则是增区间,注意“减”字