函数y=2cos²x+sin2x,x∈(0,π/2)的值域是________(求过程)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/30 07:14:17
![函数y=2cos²x+sin2x,x∈(0,π/2)的值域是________(求过程)](/uploads/image/z/6965035-43-5.jpg?t=%E5%87%BD%E6%95%B0y%3D2cos%26%23178%3Bx%2Bsin2x%2Cx%E2%88%88%280%2C%CF%80%2F2%29%E7%9A%84%E5%80%BC%E5%9F%9F%E6%98%AF________%EF%BC%88%E6%B1%82%E8%BF%87%E7%A8%8B%EF%BC%89)
函数y=2cos²x+sin2x,x∈(0,π/2)的值域是________(求过程)
函数y=2cos²x+sin2x,x∈(0,π/2)的值域是________(求过程)
函数y=2cos²x+sin2x,x∈(0,π/2)的值域是________(求过程)
y=2cos²x+sin2x
=2cos²x-1+sin2x+1
=sin2x+cos2x+1
=sin2x+sin(π/2-2x)+1
=2sinπ/4cos(2x-π/4)+1
=√2cos(2x-π/4)+1
因x∈(0,π/2),2x∈(0,π/2),2x-π/4∈(-π/4,3π/4),0
y=1+cos2x+sin2x=1+√2sin(2x+π/4)
2x+π/4=π/2,即x=π/8时,y取最大值1+√2
x=π/2时,y取最小值0.
因此值域为(0, 1+√2]
因为cos2x=2cos∧2x-1
所以2cos∧2x=cos2x+1,带入原式得
y=cos2x+sin2x+1
=√2(√2/2cos2x+√2/2sin2x)+1
=√2(sinπ/4cos2x+cosπ/4sin2x)+1
=√2sin(2x+π/4)+1
∵x∈(0,π/2),
:2x+π/4∈(π/4,5π/4)
全部展开
因为cos2x=2cos∧2x-1
所以2cos∧2x=cos2x+1,带入原式得
y=cos2x+sin2x+1
=√2(√2/2cos2x+√2/2sin2x)+1
=√2(sinπ/4cos2x+cosπ/4sin2x)+1
=√2sin(2x+π/4)+1
∵x∈(0,π/2),
:2x+π/4∈(π/4,5π/4)
所以y=√2sin(2x+π/4)最大值为√2+1,
最小值为√2sin5π/4+1=0
因此值域为(0,√2+1)
收起