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?](/uploads/image/z/9309651-51-1.jpg?t=1.%E6%B1%82y%EF%BC%9D3cos%EF%BC%88-3x-%CF%80%2F4%EF%BC%89%E5%9C%A8x%E2%88%88%EF%BC%88-%CF%80%2C%CF%80%EF%BC%89%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%3F2.cos%28%CE%B1-%CF%80%2F2%29%2Fsin%285%CF%80%2F2%2B%CE%B1%29%C3%97sin%28%CE%B1-%CF%80%29%C3%97cos%282%CF%80-%CE%B1%29%E7%AD%89%E4%BA%8E%E5%A4%9A%E5%B0%91+%E6%88%91%E5%86%99%E5%87%BA%E6%9D%A5%E8%BF%99%E9%A2%98%EF%BC%9D-sin%26%23178%3B%CE%B1%E4%B8%8D%E7%9F%A5%E9%81%93%E5%AF%B9%E4%B8%8D%E5%AF%B9%3F)
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.