Differential Calculus> On the range of a function...

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

Jack Sparrow , 12 Years ago

Grade 12

2 Answers

Revanth - -

Last Activity: 12 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].

PLEASE APPROVE IF IT IS CORRECT PLEASE

jagdish singh singh

Last Activity: 12 Years ago

$\hspace{-16}$ 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]}$$