Question

1》化简(1+sinθ+cosθ)/(sinθ/2+cosθ/2)

2》已知函数f(x)=2sin²x+sin2x,x∈[0,2π]，求使f（x）为正值的x的集合

3》已知tan2θ=-2√2，π／2＜2θ＜π，求(2cos²θ/2-sinθ-1)/(√2sin(θ+π/4)

求详细过程谢谢！！！

Answer 1

1
(1+sinθ+cosθ)/(sinθ/2+cosθ/2)
=[(1+cosθ)+sinθ]/(sinθ/2+cosθ/2)
=(2cos²θ/2+2sinθ/2cosθ/2)/(sinθ/2+cosθ/2)
=2cosθ/2(sinθ/2+cosθ/2)/(sinθ/2+cosθ/2)
=2cosθ/2
2
f(x)=2sin²x+sin2x=1-cos2x+sin2x
=√2(√2/2sin2x-√2/2cos2x)+1
=√2sin(2x-π/4)+1
由 f(x)>0 得 √2sin(2x-π/4)+1>0
即 sin(2x-π/4)>-√2/2
∴ 2kπ -π/4<2x-π/4<2kπ+5π/4,k∈Z
∴ kπ <x<kπ+3π/4,k∈Z
∴f(x)>0时，x集合为
｛x| kπ <x<kπ+3π/4,k∈Z}
3
∵tan2θ=-2√2，π／2＜2θ＜π，
∴2tanθ/(1-tan²θ)=-2√2
∴ √2tan²θ-tanθ-√2=0
∴tanθ=√2或tanθ=-√2/2
∵π／2＜2θ＜π，
∴π/4<θ<π/2
∴tanθ>0
∴tanθ=√2
∴(2cos²θ/2-sinθ-1)/(√2sin(θ+π/4)
=(cosθ-sinθ)/[√2(sinθcosπ/4+cosθsinπ/4)]
=(cosθ-sinθ)/(sinθ+cosθ)
=(1-tanθ)/(1+tanθ)
=(1-√2)/(1+√2)
=2√2-3

追问

我狠抱歉。。。我点开的时候，看见上面那位处在最上面以为是他先发的，结果是因为是推荐回答才会放在上面。。。所以抱歉了你的狠详细

追答

不怨你

Answer 2

1，(1+sinθ+cosθ)/[sin(θ/2)+cos(θ/2)]
=[2cos²(θ/2)+2sin(θ/2)cos(θ/2)]/[sin(θ/2)+cos(θ/2)]
=2cos(θ/2)[sin(θ/2)+cos(θ/2)]/[sin(θ/2)+cos(θ/2)]
=2cos(θ/2)
2，f(x)=2sin²x+sin2x=1-cos2x+sin2x=√2sin(2x-π/4)+1>0
则：sin(2x-π/4)>-√2/2
∴2kπ-π/4<2x-π/4<2kπ+5π/4
∴kπ<x<kπ+3π/4 (k∈Z)
而x∈[0,2π]
∴0<x<3π/4 ，或π<x<7π/4
3，∵tan2θ=2tanθ/(1-tan²θ)=-2√2
∴√2tan²θ-tanθ-√2=0
∴tanθ=√2，或tanθ=-√2/2
而π／2＜2θ＜π，即π／4＜θ＜π/2
∴tanθ＞0，即tanθ=√2
∴(2cos²θ/2-sinθ-1)/(√2sin(θ+π/4)
=(cosθ-sinθ)/(sinθ+cosθ)
=(1-tanθ)/(1+tanθ)
=(1-√2)/(1+√2)
=2√2-3