# What is the range of the function (arc sinx)(arc cosx)?

Revanth - -
36 Points
11 years ago

dear JACK SPARROW

its an easy question indeed . arc means nothing but the inverse function so problem becomes (sin inverse of x) (cos inverse of x) so x lies in between [-1,1].

$\hspace{-16} Let \bf{y=f(x)=\left(\sin^{-1}(x)\rigjht).\left(\cos^{-1}(x)\rigjht)}\\\\\\ Now Using \bf{*\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}\;\forall x\in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]}\\\\\\ \bf{y=\sin^{-1}(x).\left(\frac{\pi}{2}-\sin^{-1}(x)\right)}\\\\\\ Let \bf{\sin^{-1}(x)=t\Leftrightarrow t\in \left[-\frac{\pi}{2}\;,\frac{\pi}{2}\right]}\\\\\\ \bf{y=t.\left(\frac{\pi}{2}-t\right)=\frac{\pi}{2}.t-t^2}\\\\\\ \bf{y_{Max.}=-\left(t-\frac{\pi}{4}\right)^2+\frac{\pi^2}{16}\leq \frac{\pi^2}{16}}\\\\\\ Now for min...\\\\\\ \bf{-\frac{\pi}{2}\leq t \leq \frac{\pi}{2}}\\\\\\ \bf{-\frac{\pi}{2}-\frac{\pi}{4}\leq \left(t-\frac{\pi}{4}\right)\leq \frac{\pi}{2}-\frac{\pi}{4}}\\\\\\ \bf{-\frac{3\pi}{4}\leq \left(t-\frac{\pi}{4}\right)\leq \frac{\pi}{2}}\\\\\\ So \bf{y_{Min.}=\frac{\pi^2}{16}-\frac{9\pi^2}{16}\geq \frac{-\pi^2}{8}}\\\\\\ So \bf{y\in \left[\frac{-\pi^2}{8}\;,\frac{\pi^2}{16}\right]}$