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Question Number 59997 by aliesam last updated on 16/May/19

find (a,b,c) in R cos(a)+cos(b)+cos(c)=0 sin(a)+sin(b)+sin(c)=0

$${find}\:\left({a},{b},{c}\right)\:{in}\:{R}\: \\ $$$${cos}\left({a}\right)+{cos}\left({b}\right)+{cos}\left({c}\right)=\mathrm{0} \\ $$$${sin}\left({a}\right)+{sin}\left({b}\right)+{sin}\left({c}\right)=\mathrm{0} \\ $$

Answered by MJS last updated on 17/May/19

we have only 2 equations in 3 variables ⇒ we are free to choose one let a=0 1+cos b +cos c =0 sin b +sin c =0 let cos b =x∧cos c =y ⇒ sin b =±(√(1−x^2 ))∧sin c =±(√(1−y^2 )) 1+x+y=0 ⇒ y=−x−1 (√(1−x^2 ))−(√(1−y^2 ))=0 ⇒ 1−x^2 =1−y^2 1−x^2 =1−(−x−1)^2 1−x^2 =−2x−x^2 x=−(1/2) ⇒ y=−(1/2) ⇒ ⇒ cos b =cos c =−(1/2) ⇒ ⇒ b=c=±((2π)/3)+2πz∧z∈Z but sin b =−sin c ⇒ a=α∈R b=α±((2π)/3)+2πz_1 ∧c=α∓((2π)/3)+2πz_2 ∧z_1 , z_2 ∈Z

$$\mathrm{we}\:\mathrm{have}\:\mathrm{only}\:\mathrm{2}\:\mathrm{equations}\:\mathrm{in}\:\mathrm{3}\:\mathrm{variables}\:\Rightarrow \\ $$$$\mathrm{we}\:\mathrm{are}\:\mathrm{free}\:\mathrm{to}\:\mathrm{choose}\:\mathrm{one} \\ $$$$\mathrm{let}\:{a}=\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{cos}\:{b}\:+\mathrm{cos}\:{c}\:=\mathrm{0} \\ $$$$\mathrm{sin}\:{b}\:+\mathrm{sin}\:{c}\:=\mathrm{0} \\ $$$$\mathrm{let}\:\mathrm{cos}\:{b}\:={x}\wedge\mathrm{cos}\:{c}\:={y}\:\Rightarrow\:\mathrm{sin}\:{b}\:=\pm\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\wedge\mathrm{sin}\:{c}\:=\pm\sqrt{\mathrm{1}−{y}^{\mathrm{2}} } \\ $$$$\mathrm{1}+{x}+{y}=\mathrm{0}\:\Rightarrow\:{y}=−{x}−\mathrm{1} \\ $$$$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:\mathrm{1}−{x}^{\mathrm{2}} =\mathrm{1}−{y}^{\mathrm{2}} \\ $$$$\mathrm{1}−{x}^{\mathrm{2}} =\mathrm{1}−\left(−{x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}−{x}^{\mathrm{2}} =−\mathrm{2}{x}−{x}^{\mathrm{2}} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{y}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{cos}\:{b}\:=\mathrm{cos}\:{c}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\Rightarrow\:{b}={c}=\pm\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{2}\pi{z}\wedge{z}\in\mathbb{Z}\:\mathrm{but}\:\mathrm{sin}\:{b}\:=−\mathrm{sin}\:{c}\:\Rightarrow \\ $$$${a}=\alpha\in\mathbb{R} \\ $$$${b}=\alpha\pm\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{2}\pi{z}_{\mathrm{1}} \wedge{c}=\alpha\mp\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{2}\pi{z}_{\mathrm{2}} \wedge{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \in\mathbb{Z} \\ $$

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