# Is-it-possible-to-find-any-value-for-a-b-c-from-below-system-of-equetions-sina-sinb-sinc-cosa-cosb-cosc-

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