(tanx+cotx)sin2x=
答案:2 悬赏:50
解决时间 2021-11-15 02:38
- 提问者网友：枯希心
- 2021-11-14 06:14
(tanx+cotx)sin2x=
最佳答案
- 二级知识专家网友：我叫很个性
- 2021-11-14 07:42
解:
(tanx+cotx)sin2x=(sinx/cosx + cosx/sinx)sin2x=[(sin^2x+cos^2x)/(sinxcosx)]sin2x
=2[1/sin2x]sin2x=2
(tanx+cotx)sin2x=(sinx/cosx + cosx/sinx)sin2x=[(sin^2x+cos^2x)/(sinxcosx)]sin2x
=2[1/sin2x]sin2x=2
全部回答
- 1楼网友：樣嘚尐年
- 2021-11-14 08:39
f(x)=(sin2x+cos2x)/(tanx+cotx)
=(sin2x+cos2x)/(sinx/cosx+cosx/sinx)
=(sin2x+cos2x)(sinxcosx)/(sin²x+cos²x)
=(1/2)(sin2x+cos2x)sin2x
=(1/2)(sin²2x+sin2xcos2x)
=(1/4)(1-cos4x+sin4x)
=1/4+(1/4)sin4x-(1/4)cos4x
=√[(1/4)²+(-1/4)²]*sin(4x-π/4)
=1/4+√2/4*sin(4x-π/4)
(1)
周期t=2π/4=π/2
(2)
-1≤sin(4x-π/4)≤1
-√2/4≤√2/4*sin(4x-π/4≤√2/4
1/4-√2/4≤1/4+√2/4*sin(4x-π/4)≤1/4+√2/4
∴值域[(1-√2)/4，(1+√2)/4]
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