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Question Number 68414 by behi83417@gmail.com last updated on 10/Sep/19
Is it possible to find any value for a,b,c from below system of equetions? { ((sina+sinb=sinc)),((cosa+cosb=cosc)) :}
$$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{any}\:\mathrm{value}\:\mathrm{for} \\ $$$$\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}\:\mathrm{from}\:\mathrm{below}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equetions}? \\ $$$$\begin{cases}{\boldsymbol{\mathrm{sina}}+\boldsymbol{\mathrm{sinb}}=\boldsymbol{\mathrm{sinc}}}\\{\boldsymbol{\mathrm{cosa}}+\boldsymbol{\mathrm{cosb}}=\boldsymbol{\mathrm{cosc}}}\end{cases} \\ $$
Commented by kaivan.ahmadi last updated on 10/Sep/19
{ ((sinacosa+sinbcosa=sinccosa)),((−sinacosa−sinacosb=−sinacosc)) :}⇒ sinbcosa−sinacosb=sinccosa−sinacosc⇒ sin(b−a)=sin(c−a)⇒ { ((b−a=2kπ+(c−a))),((b−a=(2k+1)π+(a−c))) :} ⇒ { ((b−c=2kπ)),((b−2a+c=(2k+1)π)) :}
$$\begin{cases}{{sinacosa}+{sinbcosa}={sinccosa}}\\{−{sinacosa}−{sinacosb}=−{sinacosc}}\end{cases}\Rightarrow \\ $$$$\:\:\:\:\: \\ $$$${sinbcosa}−{sinacosb}={sinccosa}−{sinacosc}\Rightarrow \\ $$$${sin}\left({b}−{a}\right)={sin}\left({c}−{a}\right)\Rightarrow \\ $$$$\begin{cases}{{b}−{a}=\mathrm{2}{k}\pi+\left({c}−{a}\right)}\\{{b}−{a}=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi+\left({a}−{c}\right)}\end{cases} \\ $$$$ \\ $$$$\Rightarrow\begin{cases}{{b}−{c}=\mathrm{2}{k}\pi}\\{{b}−\mathrm{2}{a}+{c}=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}\end{cases} \\ $$
Answered by Tanmay chaudhury last updated on 10/Sep/19
2sin(((a+b)/2))cos(((a−b)/2))=sinc 2cos(((a+b)/2))cos(((a−b)/2))=cosc tan(((a+b)/2))=tanc ((a+b)/c)=nπ+c a+b=2nπ+2c....(1) 4sin^2 (((a+b)/2))cos^2 (((a−b)/2))=sin^2 c 4cos^2 (((a+b)/2))cos^2 (((a−b)/2))=cos^2 c add them 4cos^2 (((a−b)/2))=1 cos(((a−b)/2))=±(1/2) cosidering + sign cos(((a−b)/2))=cos((π/3)) ((a−b)/2)=2nπ+(π/3) cos(((a−b)/2))=−(1/2)=cos(π+(π/3)) (((a−b)/2))=2nπ+π+(π/3) or cos(((a−b)/2))=−(1/2)=cos(π−(π/3)) ((a−b)/2)=2nπ+((2π)/3) thus we can find value of a and b in terms of c
$$\mathrm{2}{sin}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)={sinc} \\ $$$$\mathrm{2}{cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)={cosc} \\ $$$${tan}\left(\frac{{a}+{b}}{\mathrm{2}}\right)={tanc} \\ $$$$\frac{{a}+{b}}{{c}}={n}\pi+{c} \\ $$$${a}+{b}=\mathrm{2}{n}\pi+\mathrm{2}{c}….\left(\mathrm{1}\right) \\ $$$$\mathrm{4}{sin}^{\mathrm{2}} \left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}^{\mathrm{2}} \left(\frac{{a}−{b}}{\mathrm{2}}\right)={sin}^{\mathrm{2}} {c} \\ $$$$\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}^{\mathrm{2}} \left(\frac{{a}−{b}}{\mathrm{2}}\right)={cos}^{\mathrm{2}} {c} \\ $$$${add}\:{them} \\ $$$$\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{{a}−{b}}{\mathrm{2}}\right)=\mathrm{1} \\ $$$${cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${cosidering}\:+\:{sign} \\ $$$${cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)={cos}\left(\frac{\pi}{\mathrm{3}}\right) \\ $$$$\frac{{a}−{b}}{\mathrm{2}}=\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{3}} \\ $$$${cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}={cos}\left(\pi+\frac{\pi}{\mathrm{3}}\right)\:\:\: \\ $$$$\left(\frac{\boldsymbol{{a}}−\boldsymbol{{b}}}{\mathrm{2}}\right)=\mathrm{2}\boldsymbol{{n}}\pi+\pi+\frac{\pi}{\mathrm{3}} \\ $$$${or}\:{cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}={cos}\left(\pi−\frac{\pi}{\mathrm{3}}\right) \\ $$$$\frac{{a}−{b}}{\mathrm{2}}=\mathrm{2}{n}\pi+\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\boldsymbol{{thus}}\:\boldsymbol{{we}}\:\boldsymbol{{can}}\:\boldsymbol{{find}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\boldsymbol{{a}}\:\:\boldsymbol{{and}}\:\boldsymbol{{b}}\:\boldsymbol{{in}}\:\boldsymbol{{terms}}\: \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{c}} \\ $$$$ \\ $$
Answered by mr W last updated on 10/Sep/19
sin^2 a+sin^2 b+2 sin a sin b=sin^2 c cos^2 a+cos^2 b+2 cos a cos b=cos^2 c 2+2(sin a sin b+cos a cos b)=1 sin a sin b+cos a cos b=−(1/2) cos (a−b)=−(1/2) ⇒a−b=(2n+1)π±(π/3)
$$\mathrm{sin}^{\mathrm{2}} \:{a}+\mathrm{sin}^{\mathrm{2}} \:{b}+\mathrm{2}\:\mathrm{sin}\:{a}\:\mathrm{sin}\:{b}=\mathrm{sin}^{\mathrm{2}} \:{c} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{a}+\mathrm{cos}^{\mathrm{2}} \:{b}+\mathrm{2}\:\mathrm{cos}\:{a}\:\mathrm{cos}\:{b}=\mathrm{cos}^{\mathrm{2}} \:{c} \\ $$$$\mathrm{2}+\mathrm{2}\left(\mathrm{sin}\:{a}\:\mathrm{sin}\:{b}+\mathrm{cos}\:{a}\:\mathrm{cos}\:{b}\right)=\mathrm{1} \\ $$$$\mathrm{sin}\:{a}\:\mathrm{sin}\:{b}+\mathrm{cos}\:{a}\:\mathrm{cos}\:{b}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\left({a}−{b}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{a}−{b}=\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\pm\frac{\pi}{\mathrm{3}} \\ $$

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