函数y=根号3sinx+cosx,x∈[-π/2，π/2]的值域
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解决时间 2021-02-01 02:39
- 提问者网友：世勋超人
- 2021-01-31 07:44
函数y=根号3sinx+cosx,x∈[-π/2，π/2]的值域
最佳答案
- 五星知识达人网友：思契十里
- 2021-01-31 09:14
y=2(√3/2sinx+1/2cosx)
=2(sinxcosπ/6+cosxsinπ/6)
=2sin(x+π/6)
-π/2<=x<=π/2
-π/3<=x+π/6<=2π/3
所以x+π/6=-π/3,sin(x+π/6)最小=-√3/2
x+π/6=π/2,sin(x+π/6)最大=1
所以值域[-√3,2]
=2(sinxcosπ/6+cosxsinπ/6)
=2sin(x+π/6)
-π/2<=x<=π/2
-π/3<=x+π/6<=2π/3
所以x+π/6=-π/3,sin(x+π/6)最小=-√3/2
x+π/6=π/2,sin(x+π/6)最大=1
所以值域[-√3,2]
全部回答
- 1楼网友：行雁书
- 2021-01-31 09:55
解
y=√3sinx+cosx
=2[sinx*√3/2+cosx*(1/2)]
=2[sinxcos(π/6)+cosxsin(π/6)]
=2sin(x+π/6)
x∈[0,π/2]
x+π/6∈[π/6,2π/3]
所以 sin(x+π/6)∈【1/2,1】
所以值域是[1, 2 ]
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