The domain of the function sin(arcsin(x)) is defined as x ∈ [-1, 1]. This is because arcsin(x) is only defined for x values within this interval, as the arcsine function is the inverse of the restricted sine function, which is one-to-one when limited to the domain of [-π/2, π/2]. Although sine has all real numbers as its domain, the output of arcsin(x) restricts the input for sin(arcsin(x)) to the interval [-1, 1]. Thus, the domain of sin(arcsin(x)) is directly tied to the domain of arcsin(x).
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HallsofIvy said:Since sine has all real numbers as domain, the domain of this function is the domain of arcsine. Do you understand why its domain is [-1, 1]?
The sine function is not a one-to-one function, so technically, there is no function which is the inverse of sine function.11thHeaven said:I understand why the domain of arcsin is [1,-1]. But if sine is the inverse of arcsine, shouldn't the domain of sine be the range of arcsine? I thought that was the rule where inverse functions are concerned.