Answer
$c=\pm\sqrt{1-\dfrac{4}{\pi^2}}\approx\pm 0.77118$
Work Step by Step
$y=\sin^{-1}x$ is the number in $[-\displaystyle \frac{\pi}{2},\frac{\pi}{2}]$ such that
$\sin y=x$
$f(1)=\displaystyle \frac{\pi}{2}, \quad f(-1)=-\frac{\pi}{2}$
$f(x)=\sin^{-1}x$
$f'(x)=\displaystyle \frac{1}{\sqrt{1-x^{2}}}$
f is continuous on $[a,b]=[-1,1]$, differentiable on $(a,b)=(-1,1)$, so, by the Mean Value Theorem, a $c\in(a,b)$ exists such that
$f'(c)=\displaystyle \frac{f(1)-f(-1)}{1-(-1)}$
$\displaystyle \frac{1}{\sqrt{1-c^{2}}}=\frac{\frac{\pi}{2}-(-\frac{\pi}{2})}{2}$
$\displaystyle \frac{1}{1-c^{2}}=\frac{\pi^{2}}{4}$
$1-c^{2}=\displaystyle \frac{4}{\pi^{2}}$
$c^{2}=1-\displaystyle \frac{4}{\pi^2}$
$c=\pm\sqrt{1-\frac{4}{\pi^2}}\approx\pm 0.77118$
