Question

有关三角函数问题 高手进来帮下忙

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Answer 1

1、
f(x)=4[1-cos2(π/4+x)]/2-2√3cos2x-1
=2[1-cos(π/2+2x)]-2√3cos2x-1
=2+2sin2x-2√3cos2x-1
=4(sin2x*1/2-cos2x*√3/2)+1
=4(sin2xcosπ/3-cos2xsinπ/3)+1
=4sin(2x-π/3)-1

π/4<=x<=π/2
π/6<=2x-π/3<=2π/3
所以1/2<=sin(2x-π/3)<=1
1<=4sin(2x-π/3)-1<=3
所以值域[1,3]

2、
sinx增区间是(2kπ-π/2,2kπ+π/2)
π/6<=2x-π/3<=2π/3
所以这里是(π/6,π/2)
π/6<2x-π/3<π/2
π/2<2x<5π/6
π/4<x<5π/12
同理得到减区间
所以
增区间(π/4,5π/12)
减区间(5π/12,π/2)