1.求y=3cos(-3x-π/4)在x∈(-π,π)的单调递增区间?2.cos(α-π/2)/sin(5π/2+α)×sin(α-π)×cos(2π-α)等于多少 我写出来这题=-sin²α不知道对不对?
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1.求y=3cos(-3x-π/4)在x∈(-π,π)的单调递增区间?2.cos(α-π/2)/sin(5π/2+α)×sin(α-π)×cos(2π-α)等于多少 我写出来这题=-sin²α不知道对不对?
1.求y=3cos(-3x-π/4)在x∈(-π,π)的单调递增区间?
2.cos(α-π/2)/sin(5π/2+α)×sin(α-π)×cos(2π-α)等于多少 我写出来这题=-sin²α不知道对不对?
1.求y=3cos(-3x-π/4)在x∈(-π,π)的单调递增区间?2.cos(α-π/2)/sin(5π/2+α)×sin(α-π)×cos(2π-α)等于多少 我写出来这题=-sin²α不知道对不对?
解1:
y=3cos(-3x-π/4)
y=3cos(3x+π/4)
y'=-9sin(3x+π/4)
(1)令:y'>0,即:-9sin(3x+π/4)>0
sin(3x+π/4)<0
(2k+1)π<3x+π/4<(2k+2)π
2kπ/3+π/4<x<2kπ/3+7π/12
因为:x∈(-π,π)
所以:-5π/12<x<-π/12,π/4<x<7π/12,11π/12<x<π
即:y的单调增区间是:x∈(-5π/12,-π/12)∪(π/4,7π/12)∪(11π/12,π)
(2)同理,令:y'<0,有:sin(3x+π/4)>0
解得:y的单调减区间是:x∈(-7π/12,-5π/12)∪(-π/12,3π/12)∪(7π/12,11π/12)
综上所述:
y的单调增区间是:x∈(-5π/12,-π/12)∪(π/4,7π/12)∪(11π/12,π)
y的单调减区间是:x∈(-7π/12,-5π/12)∪(-π/12,π/4)∪(7π/12,11π/12)
解2:
cos(α-π/2)/sin(5π/2+α)×sin(α-π)×cos(2π-α)
=-cos(π/2-α)/sin(2π+π/2+α)×sin(π-α)×cos(2π-α)
=-sinα/cosα×sinα×cosα
=-(sinα)^2
楼主做的是对的.