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Question Number 137038 by bobhans last updated on 29/Mar/21

$$$$ Given triangle ABC, what is the maximum value of y=2cosA + cosB + cosC?\n

Answered by mr W last updated on 29/Mar/21

$$A=\pi -(B+C)$$ $$y=-2\cos (B+C)+\cos B+\cos C$$ $$duetosymmetry:B=C=x$$ $$y=-2\cos 2x+2\cos x$$ $$y=-4{\cos }^{2}x+2+2\cos x)$$ $$y=\frac{9}{4}-(4{\cos }^{2}x-2\cos x+\frac{1}{4})$$ $$y=\frac{9}{4}-{(2\cos x-\frac{1}{2})}^2$$ $$y_{max}=\frac{9}{4}whenB=C=x={\cos }^{-1}\frac{1}{4}$$

Commented bybobhans last updated on 29/Mar/21

$$\mathrm{why}\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{symetri}\mathrm{sir}$$

Commented bymr W last updated on 29/Mar/21

$$inthefunction$$ $$y=-2\cos (B+C)+\cos B+\cos C$$ $$youcanexchangeBandCandthe$$ $$functionremainsthesame.when$$ $$suchafunctionhasmaximumor$$ $$minimum,thenB=C.$$

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