The sides 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} ) and cos( \gamma + \frac{\pi}{2} ) is equal to..............

Question

The sides of a triangle inscribed in a given circle subtend anglesn alpha,beta and gamma at the centre. The minimum value ofn the arithmetic mean ofn cos( alpha + (pi / 2) ),cos( beta + (pi / 2) )n and cos( gamma + (pi / 2) ) is equaln to..............

DivineJEE · Accepted Answer

We have alpha+beta+gamma=2 pinnnn => (alpha / 2)+(beta / 2)+(gamma / 2)=pinnn nArithmetic meannnnnn n =(1 / 3)[cos (alpha+(pi / 2))+cos (beta+(pi / 2))+cos (gamma+(pi / 2))] n =(-1 / 3)[sin alpha+sin beta+sin gamma] n =(-1 / 3)[2 sin (alpha+beta / 2) cos (alpha-beta / 2)+2 sin (gamma / 2) cos (gamma / 2)] n =(-1 / 3)[2 sin (pi-(gamma / 2)) cos ((alpha-beta / 2))+2 sin (gamma / 2) cos (gamma / 2)] n =(-1 / 3) 2 sin (gamma / 2)[cos ((alpha-beta / 2))+cos (pi-(alpha+beta / 2))] n =(-1 / 3) x 2 sin (gamma / 2)[cos (alpha-beta / 2)-cos (alpha+beta / 2)] n =(-1 / 3) x 2 sin (gamma / 2) 2 sin (alpha / 2) sin (beta / 2)=(-4 / 3) sin (alpha / 2) sin (beta / 2) sin (gamma / 2)nn nnnn∴ The arithmetic mean will be least when sin ((alpha / 2)),nnsin ((beta / 2)), sin ((gamma / 2)) is greatest and sin ((alpha / 2)) sin ((beta / 2)) sin ((gamma / 2)) is greatest when sin ((alpha / 2))= sin ((beta / 2))=sin ((gamma / 2)),nnnnalpha=beta=gamma=120^°nnn nThus minimum value of the arithmetic meannnnn=(-4 / 3) sin ^3 60^°=(-4 / 3) x frac3 sqrt(3)8=-fracsqrt(3)2nnn

Explanation

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