求函数y=2sin(2x+π/3),x属于(-π/6,π/6)的值域
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求函数y=2sin(2x+π/3),x属于(-π/6,π/6)的值域
要过程
要过程
![求函数y=2sin(2x+π/3),x属于(-π/6,π/6)的值域](/uploads/image/z/4453602-42-2.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0y%3D2sin%282x%2B%CF%80%2F3%29%2Cx%E5%B1%9E%E4%BA%8E%28-%CF%80%2F6%2C%CF%80%2F6%29%E7%9A%84%E5%80%BC%E5%9F%9F)
x∈(-π/6,π/6)
2x∈(-π/3,π/3)
2x+π/3∈(0,2π/3)
sin(2x+π/3)∈(0,1]
2sin(2x+π/3)∈(0,2]
所以函数y=2sin(2x+π/3),x属于(-π/6,π/6)的值域为:(0,2]
再问: 为什么sin2π/3=1?
再答: 2x+π/3∈(0,2π/3) 当2x+π/3=π/2时 sin(2x+π/3)=1
2x∈(-π/3,π/3)
2x+π/3∈(0,2π/3)
sin(2x+π/3)∈(0,1]
2sin(2x+π/3)∈(0,2]
所以函数y=2sin(2x+π/3),x属于(-π/6,π/6)的值域为:(0,2]
再问: 为什么sin2π/3=1?
再答: 2x+π/3∈(0,2π/3) 当2x+π/3=π/2时 sin(2x+π/3)=1
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