Question Number 137038 by bobhans last updated on 29/Mar/21
$$ \\ $$Given triangle ABC, what is the maximum value of y=2cosA + cosB + cosC?
Answered by mr W last updated on 29/Mar/21
$${A}=\pi−\left({B}+{C}\right) \\ $$$${y}=−\mathrm{2}\:\mathrm{cos}\:\left({B}+{C}\right)+\mathrm{cos}\:{B}+\mathrm{cos}\:{C} \\ $$$${due}\:{to}\:{symmetry}:\:{B}={C}={x} \\ $$$${y}=−\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{x}+\mathrm{2}\:\mathrm{cos}\:{x} \\ $$$$\left.{y}=−\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{x}+\mathrm{2}+\mathrm{2}\:\mathrm{cos}\:{x}\right) \\ $$$${y}=\frac{\mathrm{9}}{\mathrm{4}}−\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{2}\:\mathrm{cos}\:{x}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${y}=\frac{\mathrm{9}}{\mathrm{4}}−\left(\mathrm{2}\:\mathrm{cos}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${y}_{{max}} =\frac{\mathrm{9}}{\mathrm{4}}\:{when}\:{B}={C}={x}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by bobhans last updated on 29/Mar/21
$$\mathrm{why}\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{symetri}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 29/Mar/21
$${in}\:{the}\:{function} \\ $$$${y}=−\mathrm{2}\:\mathrm{cos}\:\left({B}+{C}\right)+\mathrm{cos}\:{B}+\mathrm{cos}\:{C} \\ $$$${you}\:{can}\:{exchange}\:{B}\:{and}\:{C}\:{and}\:{the} \\ $$$${function}\:{remains}\:{the}\:{same}.\:{when}\: \\ $$$${such}\:{a}\:{function}\:{has}\:{maximum}\:{or}\: \\ $$$${minimum},\:{then}\:{B}={C}. \\ $$