Which trigonometric function requires a domain restriction of [0,π/2)u(π/2,π] to make it invertable?
The trigonometric function that requires a domain restriction of [0,π/2) ∪ (π/2,π] to make it invertible is the tangent function (tan(x)).
The trigonometric function that requires a domain restriction of [0,π/2) ∪ (π/2,π] to make it invertible is the tangent function, often denoted as tan(x).
To understand why the tangent function needs this domain restriction, let's first discuss what it means for a function to be invertible. An invertible function is a function where each element in the range is paired with exactly one element in the domain. In other words, every output value has a unique input value associated with it.
The tangent function is defined as the ratio of the sine and cosine functions: tan(x) = sin(x) / cos(x). However, the cosine function has zeros at x = π/2 and x = 3π/2, which means the tangent function would be undefined at those points.
If we consider the full domain of tangent (−∞, +∞), it becomes clear that at every multiple of π/2, the tangent function has vertical asymptotes, meaning the function approaches positive or negative infinity as x approaches those values.
To make the tangent function invertible, we need to restrict its domain to avoid these vertical asymptotes. By excluding the points where the tangent function is undefined or has vertical asymptotes, we limit the domain to [0,π/2) ∪ (π/2,π], which allows each x-value in the range to have a unique corresponding tangent value.
In summary, the domain restriction of [0,π/2) ∪ (π/2,π] makes the tangent function invertible because it avoids the vertical asymptotes at x = π/2 and x = 3π/2 where the function is undefined and loses its one-to-one nature.
looks good to me.
A quick check in your trig book will confirm it ...