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Question-162344




Question Number 162344 by stelor last updated on 28/Dec/21
Commented by mr W last updated on 29/Dec/21
two equations for three unknowns  ⇒no unique solution!
$${two}\:{equations}\:{for}\:{three}\:{unknowns} \\ $$$$\Rightarrow{no}\:{unique}\:{solution}! \\ $$
Answered by aleks041103 last updated on 29/Dec/21
cosa+cosb+cosc = 0 (1)  sina+sinb+sinc = 0 (2)  ⇒e^(ia) +e^(ib) +e^(ic) =0  ⇒e^(i(a−c)) +e^(i(b−c)) +1=0  ⇒e^(i(a−c)) =(e^(i(b−c)) )^∗ =e^(i(c−b))   a−c=c−b  a+b=2c  also  −(1/2)=cos(a−c)  ⇒a−c=((2π)/3) and ((4π)/3)  ⇒a=c±((2π)/3)  ⇒b=c±(−((2π)/3))
$${cosa}+{cosb}+{cosc}\:=\:\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$${sina}+{sinb}+{sinc}\:=\:\mathrm{0}\:\left(\mathrm{2}\right) \\ $$$$\Rightarrow{e}^{{ia}} +{e}^{{ib}} +{e}^{{ic}} =\mathrm{0} \\ $$$$\Rightarrow{e}^{{i}\left({a}−{c}\right)} +{e}^{{i}\left({b}−{c}\right)} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{e}^{{i}\left({a}−{c}\right)} =\left({e}^{{i}\left({b}−{c}\right)} \right)^{\ast} ={e}^{{i}\left({c}−{b}\right)} \\ $$$${a}−{c}={c}−{b} \\ $$$${a}+{b}=\mathrm{2}{c} \\ $$$${also} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}={cos}\left({a}−{c}\right) \\ $$$$\Rightarrow{a}−{c}=\frac{\mathrm{2}\pi}{\mathrm{3}}\:{and}\:\frac{\mathrm{4}\pi}{\mathrm{3}} \\ $$$$\Rightarrow{a}={c}\pm\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\Rightarrow{b}={c}\pm\left(−\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$
Commented by aleks041103 last updated on 29/Dec/21
this has a cute geometric proof too.  the complex vectors e^(ia) , e^(ib) , e^(ic)  must  sum up to 0 and have the same length.  Therefore the tree vectors must be symmetric  ⇒ the angle between any two of them  must be ((360°)/3)=120°.
$${this}\:{has}\:{a}\:{cute}\:{geometric}\:{proof}\:{too}. \\ $$$${the}\:{complex}\:{vectors}\:{e}^{{ia}} ,\:{e}^{{ib}} ,\:{e}^{{ic}} \:{must} \\ $$$${sum}\:{up}\:{to}\:\mathrm{0}\:{and}\:{have}\:{the}\:{same}\:{length}. \\ $$$${Therefore}\:{the}\:{tree}\:{vectors}\:{must}\:{be}\:{symmetric} \\ $$$$\Rightarrow\:{the}\:{angle}\:{between}\:{any}\:{two}\:{of}\:{them} \\ $$$${must}\:{be}\:\frac{\mathrm{360}°}{\mathrm{3}}=\mathrm{120}°. \\ $$

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