Let f(x)=((^-1x)+(^-1x))^2-1, |x|1.

Question

Let $f(x)=(\sin(	an^{-1}x)+\sin(\cot^{-1}x))^2-1$, $|x|>1$. If $\dfrac{dy}{dx}=\dfrac12\dfrac{d}{dx}(\sin^{-1}(f(x)))$ and $y(\sqrt3)=\dfrac{\pi}{6}$, then $y(-\sqrt3)$ equals:

Accepted Answer

$f(x)=\sin(2	an^{-1}x)$, so $2y=\sin^{-1}(\sin(2	an^{-1}x))+C$. Using $x=\sqrt3$: $\dfrac{\pi}{3}=\dfrac{\pi}{3}+C \implies C=0$.

At $x=-\sqrt3$: $2y=\sin^{-1}(\sin(-2\pi/3)) = -\dfrac{\pi}{3} \implies y=-\dfrac{\pi}{6}$.

Solutions · 1