Question

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

• A. $-\frac{\sqrt{3}}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. None of these

Answer 1

The correct option is A $-\frac{\sqrt{3}}{2}$

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

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

$=-\frac{1}{3}[\sin \alpha +\sin \beta +\sin \gamma ]=-\frac{4}{3}\sin \left(\frac{\alpha }{2}\right)\sin \left(\frac{\beta }{2}\right)\sin \left(\frac{\gamma }{2}\right)$

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

$A.M.$ will be least if $\sin \left(\frac{\alpha }{2}\right)\sin \left(\frac{\beta }{2}\right)\sin \left(\frac{\gamma }{2}\right)$ is greatest i.e $\sin A\sin B\sin C$ is greatest,

We know that in $△ABC,\sin A\sin B\sin C$ is greatest if $A=B=C=\frac{\pi }{3}$

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