发布网友 发布时间:2024-10-23 17:06
共1个回答
热心网友 时间:2024-10-31 09:52
sin(π-α)=4/5; α∈(0,π/2)
sin(π-α)=sinα=4/5
cosα=3/5
sin2α=2*sinα*cosα=24/25
(2)
f(x)=5/3cosαsin2x-cos2x
=(5/3)*(3/5)sin2x-cos2x
=sin2x-cos2x
=√2sin(2x-π/4)
将2x-π/4代入到标准的正弦中去解出x即为所求,
-π/2+2kπ≤2x-π/4≤π/2+2kπ
-π/8+kπ≤x≤3π/8+kπ
所以函数的单调增区间为:
【-π/8+kπ,3π/8+kπ】