Question

三角函数y=sin(2x-π/6),求出其值域,单调区间,对称轴,以及由y=sin x经过怎样的变化得到的

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Answer 1

y=sin(2x-π/6),其值域, [-1,1]
单调区间
2kπ-π/2<=2x-π/6<=2kπ+π/2 kπ-π/6<=x<=kπ+π/3
增区间【kπ-π/6，kπ+π/3】k∈Z
2kπ+π/2<=2x-π/6<=2kπ+3π/2 kπ+π/3<=x<=kπ+5π/6
减区间【kπ+π/3，kπ+5π/6】k∈Z
对称轴 2x-π/6=kπ+π/2 x=kπ+π/3 k∈Z
y=sinx纵坐标不变，横坐标缩短为原来的一半，得到y=sin2x
再向右平移π/12个单位得到 y=sin(2x-π/6),