Inverse Trigonometric Functions

We have also learnt in Chapter 1 that if f: X→Y such that \(f(x)=y\) is one-one and onto, then we can define a unique function \(g: Y \rightarrow X\) such that \(g(y)=x\), where \(x \in X\) and \(y=f(x)\), \(y \in Y\). Here, the domain of g = range of f and the range of g = domain of f. The function g is called the inverse of f and is denoted by \(f^{-1}\). Further, g is also one-one and onto and inverse of g is f. Thus, \(g^{-1} = (f^{-1})^{-1} = f\). We also have \((f^{-1} \circ f)(x) = f^{-1}(f(x)) = f^{-1}(y) = x\) and \((f \circ f^{-1})(y) = f(f^{-1}(y)) = f(x) = y\).

Since the domain of sine function is the set of all real numbers and range is the closed interval [-1, 1]. If we restrict its domain to \([\frac{-\pi}{2}, \frac{\pi}{2}]\) then it becomes one-one and onto with range [-1, 1]. Actually, sine function restricted to any of the intervals \([\frac{-3\pi}{2}, \frac{-\pi}{2}], [\frac{-\pi}{2}, \frac{\pi}{2}], [\frac{\pi}{2}, \frac{3\pi}{2}]\) etc., is one-one and its range is [-1, 1]. We can, therefore, define the inverse of sine function in each of these intervals. We denote the inverse of sine function by \(sin^{-1}\) (arc sine function). Thus, \(sin^{-1}\) is a function whose domain is [-1, 1] and range could be any of the intervals \([\frac{-3\pi}{2}, \frac{-\pi}{2}], [\frac{-\pi}{2}, \frac{\pi}{2}]\) or \([\frac{\pi}{2}, \frac{3\pi}{2}]\) and so on. Corresponding to each such interval, we get a branch of the function \(sin^{-1}\). The branch with range \([\frac{-\pi}{2}, \frac{\pi}{2}]\) is called the principal value branch, whereas other intervals as range give different branches of \(sin^{-1}\). When we refer to the function \(sin^{-1}\), we take it as the function whose domain is [-1, 1] and range is \([\frac{-\pi}{2}, \frac{\pi}{2}]\). We write \(sin^{-1}: [-1, 1] \rightarrow [\frac{-\pi}{2}, \frac{\pi}{2}]\) From the definition of the inverse functions, it follows that sin\((sin^{-1}x) = x\) if \(-1 \le x \le 1\) and \(sin^{-1}(sin x) = x\) if \(-\frac{\pi}{2} \le x \le \frac{\pi}{2}\). In other words, if \(y = sin^{-1}x\), then sin \(y = x\).