Question

What is the domain of the function sin^-1 (2x)?​

Answer 1

$\Large\bold{\underline{\underline{Solution:-}}}$

$let \\ y = {sin}^{ - 1} (2x) \\ 2x = \sin(y) \\ x = \frac{ \sin(y) }{2} \\ \\ domain \: of \: {sin}^{ - 1} (2x) \: is \: similarly \: \\ to \: the \: range \: of \: \frac{ \sin(y) }{2} \\ \\ x = \frac{ \sin(y) }{2} \\ \\ as \: we \: know \: that \: sin(y) \: is \: always \\ lie \: between \: - 1 \: to \: + 1 \\ (including \: - 1 \: and \: + 1) \\ so \: x \: will \: always \: lie \: between \: \\ - \frac{1}{2} \: to \: \frac{1}{2} \: which \: is \: the \: required \\ \: answer$