The sides of a triangle inscribed in a given circle subtend angles $\alpha ,\beta ,\gamma$ at the center. The minimum value of the $A.M.$ of $\cos \left(\alpha +\frac{\pi }{2}\right),\cos \left(\beta +\frac{\pi }{2}\right)$ and $\cos \left(\gamma +\frac{\pi }{2}\right)$ is equal to

Let the triangle is $ABC$,

Since the angle made by a chord at the center of a circle is twice the angle made by the chord at the circumference, we get

$\angle A=\frac{\alpha }{2},\angle B=\frac{\beta }{2} ,\angle C=\frac{\gamma }{2}\Rightarrow \alpha +\beta +\gamma =2\pi$

$A.M.=\frac{1}{3}\left[\cos \right(\alpha +\frac{\mathrm{\pi }}{2} )+\cos (\beta +\frac{\mathrm{\pi }}{2} )+\cos (\gamma +\frac{\mathrm{\pi }}{2})]$

$=- \frac{1}{3} [\mathrm{sin\alpha }+\mathrm{sin\beta }+\mathrm{sin\gamma }]$

$=-\frac{4}{3}\sin ( \frac{\mathrm{\alpha }}{2} )\sin ( \frac{\mathrm{\beta }}{2})\sin ( \frac{\mathrm{\gamma }}{2})$

$=-\frac{4}{3}\sin A\sin B\sin C$

$A.M.$ is least if $\sin \left(A\right)\sin \left(B\right)\sin \left(C\right)$ is maximum

For this $A=B=C=\frac{\pi }{3}$

$\Rightarrow \text{Least} A.M. = \frac{-4}{3}{\left(\frac{\sqrt{3}}{2}\right)}^3=-\frac{\sqrt{3}}{2}$