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Question Number 185486 by sonukgindia last updated on 22/Jan/23

Answered by a.lgnaoui last updated on 23/Jan/23

A−(B+C)=0 A+B+C=2(B+C)=2A A+B−C=A−(B+C)+2B=2B A−B+C=A−(B+C)+2C=2C ((sin (A+B+C)+sin (A+B−C)+sin (A−B+C))/(2sin Acos Bcos C))= ((sin 2A+sin 2B+sin 2C)/(2sin Acos Bcos C))= ((2[sin Acos A+sin Bcos B+sin Ccos C])/(2sin Acos Bcos C))= ((cos A)/(cos Bcos C))+((tan B)/(sin Acos C))+((tan C)/(sin Acos B)) =((cos A)/(cos Bcos C))+(1/(sin A))(((tan B)/(cos C))+((tan C)/(cos B))) =((cos A)/(cos Bcos C))+(1/(sin A))(((sin B+sin C)/(cos Bcos C))) =(1/(cos Bcos C))[cos A+((sin B+sin C)/(sin A))] (1) sin A=sin (B+C) (1)=(1/(cos B cos C))[(cos (B+C)+((sin B+sin C)/(sin (B+C))))] =(1/(cos A+sinBsinC ))[cos A+((sin B+sin C)/(sin A))]

$$\mathrm{A}−\left(\mathrm{B}+\mathrm{C}\right)=\mathrm{0} \\ $$$$\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{2}\left(\mathrm{B}+\mathrm{C}\right)=\mathrm{2A} \\ $$$$\mathrm{A}+\mathrm{B}−\mathrm{C}=\mathrm{A}−\left(\mathrm{B}+\mathrm{C}\right)+\mathrm{2B}=\mathrm{2B} \\ $$$$\mathrm{A}−\mathrm{B}+\mathrm{C}=\mathrm{A}−\left(\mathrm{B}+\mathrm{C}\right)+\mathrm{2C}=\mathrm{2C} \\ $$$$ \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)+\mathrm{sin}\:\left(\mathrm{A}+\mathrm{B}−\mathrm{C}\right)+\mathrm{sin}\:\left(\mathrm{A}−\mathrm{B}+\mathrm{C}\right)}{\mathrm{2sin}\:\mathrm{Acos}\:\mathrm{Bcos}\:\mathrm{C}}= \\ $$$$\frac{\mathrm{sin}\:\mathrm{2A}+\mathrm{sin}\:\mathrm{2B}+\mathrm{sin}\:\mathrm{2C}}{\mathrm{2sin}\:\mathrm{Acos}\:\mathrm{Bcos}\:\mathrm{C}}= \\ $$$$\frac{\mathrm{2}\left[\mathrm{sin}\:\mathrm{Acos}\:\mathrm{A}+\mathrm{sin}\:\mathrm{Bcos}\:\mathrm{B}+\mathrm{sin}\:\mathrm{Ccos}\:\mathrm{C}\right]}{\mathrm{2sin}\:\mathrm{Acos}\:\mathrm{Bcos}\:\mathrm{C}}= \\ $$$$\frac{\mathrm{cos}\:\mathrm{A}}{\mathrm{cos}\:\mathrm{Bcos}\:\mathrm{C}}+\frac{\mathrm{tan}\:\mathrm{B}}{\mathrm{sin}\:\mathrm{Acos}\:\mathrm{C}}+\frac{\mathrm{tan}\:\mathrm{C}}{\mathrm{sin}\:\mathrm{Acos}\:\mathrm{B}} \\ $$$$=\frac{\mathrm{cos}\:\mathrm{A}}{\mathrm{cos}\:\mathrm{Bcos}\:\mathrm{C}}+\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{A}}\left(\frac{\mathrm{tan}\:\mathrm{B}}{\mathrm{cos}\:\mathrm{C}}+\frac{\mathrm{tan}\:\mathrm{C}}{\mathrm{cos}\:\mathrm{B}}\right) \\ $$$$=\frac{\mathrm{cos}\:\mathrm{A}}{\mathrm{cos}\:\mathrm{Bcos}\:\mathrm{C}}+\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{A}}\left(\frac{\mathrm{sin}\:\mathrm{B}+\mathrm{sin}\:\mathrm{C}}{\mathrm{cos}\:\mathrm{Bcos}\:\mathrm{C}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{Bcos}\:\mathrm{C}}\left[\mathrm{cos}\:\mathrm{A}+\frac{\mathrm{sin}\:\mathrm{B}+\mathrm{sin}\:\mathrm{C}}{\mathrm{sin}\:\mathrm{A}}\right]\:\left(\mathrm{1}\right) \\ $$$$\mathrm{sin}\:\mathrm{A}=\mathrm{sin}\:\left(\mathrm{B}+\mathrm{C}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{B}}\:\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{C}}}\left[\left(\boldsymbol{\mathrm{cos}}\:\left(\boldsymbol{\mathrm{B}}+\boldsymbol{\mathrm{C}}\right)+\frac{\boldsymbol{\mathrm{sin}}\:\mathrm{B}+\boldsymbol{\mathrm{sin}}\:\mathrm{C}}{\boldsymbol{\mathrm{sin}}\:\left(\boldsymbol{\mathrm{B}}+\boldsymbol{\mathrm{C}}\right)}\right)\right] \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{sinBsinC}}\:}\left[\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{A}}+\frac{\boldsymbol{\mathrm{sin}}\:\boldsymbol{\mathrm{B}}+\boldsymbol{\mathrm{sin}}\:\boldsymbol{\mathrm{C}}}{\boldsymbol{\mathrm{sin}}\:\boldsymbol{\mathrm{A}}}\right] \\ $$

Commented by mr W last updated on 23/Jan/23

this is like when you try to simplify (1/2) to (4/(2+(3/(1−(1/2))))).

$${this}\:{is}\:{like}\:{when}\:{you}\:{try}\:{to}\:{simplify}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:{to}\:\frac{\mathrm{4}}{\mathrm{2}+\frac{\mathrm{3}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}}. \\ $$

Answered by Frix last updated on 23/Jan/23

a=b+c sin (a+b+c) +sin (a+b−c) +sin (a−b+c) = =sin (2b+2c) +sin 2b +sin 2c 2sin a cos b cos c = =2cos^2 b sin c cos c +2sin b cos b cos^2 c = =((sin (2b+2c) +sin 2b +sin 2c)/2) ⇒ answer is 2

$${a}={b}+{c} \\ $$$$\mathrm{sin}\:\left({a}+{b}+{c}\right)\:+\mathrm{sin}\:\left({a}+{b}−{c}\right)\:+\mathrm{sin}\:\left({a}−{b}+{c}\right)\:= \\ $$$$=\mathrm{sin}\:\left(\mathrm{2}{b}+\mathrm{2}{c}\right)\:+\mathrm{sin}\:\mathrm{2}{b}\:+\mathrm{sin}\:\mathrm{2}{c} \\ $$$$\mathrm{2sin}\:{a}\:\mathrm{cos}\:{b}\:\mathrm{cos}\:{c}\:= \\ $$$$=\mathrm{2cos}^{\mathrm{2}} \:{b}\:\mathrm{sin}\:{c}\:\mathrm{cos}\:{c}\:+\mathrm{2sin}\:{b}\:\mathrm{cos}\:{b}\:\mathrm{cos}^{\mathrm{2}} \:{c}\:= \\ $$$$=\frac{\mathrm{sin}\:\left(\mathrm{2}{b}+\mathrm{2}{c}\right)\:+\mathrm{sin}\:\mathrm{2}{b}\:+\mathrm{sin}\:\mathrm{2}{c}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\mathrm{answer}\:\mathrm{is}\:\mathrm{2} \\ $$

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