Question and Answers Forum

All Question   Topic List

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=π−(B+C) y=−2 cos (B+C)+cos B+cos C due to symmetry: B=C=x y=−2 cos 2x+2 cos x y=−4 cos^2 x+2+2 cos x) y=(9/4)−(4 cos^2 x−2 cos x+(1/4)) y=(9/4)−(2 cos x−(1/2))^2 y_(max) =(9/4) when B=C=x=cos^(−1) (1/4)

$${A}=\pi−\left({B}+{C}\right) \\ $$ $${y}=−\mathrm{2}\:\mathrm{cos}\:\left({B}+{C}\right)+\mathrm{cos}\:{B}+\mathrm{cos}\:{C} \\ $$ $${due}\:{to}\:{symmetry}:\:{B}={C}={x} \\ $$ $${y}=−\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{x}+\mathrm{2}\:\mathrm{cos}\:{x} \\ $$ $$\left.{y}=−\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{x}+\mathrm{2}+\mathrm{2}\:\mathrm{cos}\:{x}\right) \\ $$ $${y}=\frac{\mathrm{9}}{\mathrm{4}}−\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{2}\:\mathrm{cos}\:{x}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$ $${y}=\frac{\mathrm{9}}{\mathrm{4}}−\left(\mathrm{2}\:\mathrm{cos}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$ $${y}_{{max}} =\frac{\mathrm{9}}{\mathrm{4}}\:{when}\:{B}={C}={x}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented bybobhans last updated on 29/Mar/21

why B and C symetri sir

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

Commented bymr W last updated on 29/Mar/21

in the function y=−2 cos (B+C)+cos B+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.

$${in}\:{the}\:{function} \\ $$ $${y}=−\mathrm{2}\:\mathrm{cos}\:\left({B}+{C}\right)+\mathrm{cos}\:{B}+\mathrm{cos}\:{C} \\ $$ $${you}\:{can}\:{exchange}\:{B}\:{and}\:{C}\:{and}\:{the} \\ $$ $${function}\:{remains}\:{the}\:{same}.\:{when}\: \\ $$ $${such}\:{a}\:{function}\:{has}\:{maximum}\:{or}\: \\ $$ $${minimum},\:{then}\:{B}={C}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com