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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The sides of a triangle inscribed in a given circle .Subtend angles of $\alpha,\beta$ and $\gamma$ at the centre.The minimum value of the arithemetic mean of $\cos(\alpha+\pi/2),\cos(\beta+\pi/2)$ and $\cos(\gamma+\pi/2)$ is equal to

$(a)\;-\large\frac{\sqrt 3}{2}\qquad$$(b)\;\large\frac{1}{\sqrt 2}$$\qquad(c)\;\large\frac{2}{3}$$\qquad(d)\;\large\frac{3}{2}$

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We know that A.M $\geq$ G.M
$\Rightarrow$ Min value of AM is obtained when AM=GM
$\Rightarrow$The quantities whose AM is being taken are equal.
(i.e) $\cos(\alpha+\pi/2)=\cos(\beta+\alpha/2)=\cos(\gamma+\pi/2)$
Also $\alpha+\beta+\gamma=360^{\large\circ}$
$\Rightarrow \alpha=\beta=\gamma=120^{\large\circ}$
$\therefore$ Minimum value of A.M=$\large\frac{\cos\big(\Large\frac{2\pi}{3}+\frac{\pi}{2}\big)+\cos\big(\Large\frac{2\pi}{3}+\frac{\pi}{3}\big)+\cos\big(\Large\frac{2\pi}{3}+\frac{\pi}{2}\big)}{3}$
$\Rightarrow \large\frac{-\sin\Large\frac{2\pi}{3}}{3}$
$\Rightarrow -\large\frac{\sqrt 3}{2}$
Hence (a) is the correct answer.
answered Oct 14, 2013 by sreemathi.v

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