〔sinα-cosα)^2=1-2sinαcosα=2-m^2,当α∈{2kπ+π/4,2kπ+5π/4}(k∈z)时,sinα-cosα≥0,
原式=√2-m^2.这个范围怎么求!
人气:450 ℃ 时间:2020-05-01 11:08:16
解答
〔sinα-cosα)^2
=1-2sinαcosα
=1-sin2α
α∈{2kπ+π/4,2kπ+5π/4}(k∈z).则2α∈R
1-sin2α∈[0,2]
〔sinα-cosα)^2=1-sin2α=√2-m^2∈[0,2]
然后就简单了,自己算啦
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