Question

The side of a triangle inscribed in a given circle subtend angles, $\alpha ,\beta$ and $\gamma$ at the centre. The minimum value of the arithmetic mean of $\cos [\alpha +\frac{\pi }{2}],\cos [\beta +\frac{\pi }{2}]\text{and}\cos [\gamma +\frac{\pi }{2}]$

Answer 1

The arithmetic mean of $cos(\alpha +\frac{\pi }{2})$, $cos(\beta +\frac{\pi }{2})$ and $cos(\gamma +\frac{\pi }{2})$ is,

$A.M.=\frac{cos(\alpha +\frac{\pi }{2})+cos(\beta +\frac{\pi }{2})+cos(\gamma +\frac{\pi }{2})}{3}$

We know that, $A.M.\ge G.M.$

For A.M. to be minimum, $A.M.=G.M.$

To satisfy this condition,

$(\alpha +\frac{\pi }{2})=(\beta +\frac{\pi }{2})=(\gamma +\frac{\pi }{2})$

$\therefore -sin\alpha =-sin\beta =-sin\gamma$

$\therefore sin\alpha =sin\beta =sin\gamma$

$\therefore \alpha =\beta =\gamma$ (1)

From the figure,

$\alpha +\beta +\gamma =2\pi$

$\therefore \alpha +\alpha +\alpha =2\pi$

$\therefore 3\alpha =2\pi$

$\therefore \alpha =\frac{2\pi }{3}$

Thus, minimum value of A.M. is,

$A.M.=\frac{-3sin\left(\frac{2\pi }{3}\right)}{3}$

$\therefore A.M.=-sin(\pi -\frac{\pi }{3})$

$\therefore A.M.=-sin\frac{\pi }{3}$

$\therefore A.M.=-\frac{\sqrt{3}}{2}$