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Question Number 209560 by MM42 last updated on 15/Jul/24
In the triangle ABC ; cos(B−C)=(1/3) Show that : ((1−3cos(B+C))/(6sinBcosC))=tanC
$${In}\:{the}\:{triangle}\:{ABC}\:;\:{cos}\left({B}−{C}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${Show}\:{that}\::\:\:\frac{\mathrm{1}−\mathrm{3}{cos}\left({B}+{C}\right)}{\mathrm{6}{sinBcosC}}={tanC} \\ $$$$ \\ $$
Answered by Spillover last updated on 14/Jul/24
((1−3×(1/3))/(6sinBcosC))=tanC 0=tan C ...... something wrong
$$\:\:\frac{\mathrm{1}−\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{6}{sinBcosC}}={tanC} \\ $$$$\mathrm{0}=\mathrm{tan}\:{C} \\ $$$$…… \\ $$$${something}\:{wrong} \\ $$
Commented by MM42 last updated on 15/Jul/24
corrected
$${corrected} \\ $$
Answered by Berbere last updated on 15/Jul/24
cos(B−C)=(1/3) Without lost of Generality B>C B−C=arcos((1/3)) A+B+C=π B+C=arcos((1/3))+2C sin(B)cos(c)=sin(c+cos^(−1) ((1/3))).cos(c) ((1/3)sin(c)+((2(√2))/3)cos(c))cosC) 1−3cos(cos^(−1) ((1/3))+2c)=1−cos(2c)+3.((2(√2))/3)sin(2c) =2sin^2 (c)−4(√2)sin(c)cos(c)=sin(c)(2sin(c)+4(√2)cos(c)) ((1−3cos(B+C)))/(6sin(B)cos(C)))=((sin(C)(2sin(C)+4(√2)cos(C)))/((2sin(C)+4(√2)cos(C))cos(C)))=tan (C)
$${cos}\left({B}−{C}\right)=\frac{\mathrm{1}}{\mathrm{3}}\:{Without}\:{lost}\:{of}\:{Generality}\:{B}>{C} \\ $$$${B}−{C}={arcos}\left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${A}+{B}+{C}=\pi \\ $$$${B}+{C}={arcos}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{2}{C} \\ $$$${sin}\left({B}\right){cos}\left({c}\right)={sin}\left({c}+\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right).{cos}\left({c}\right) \\ $$$$\left.\left(\frac{\mathrm{1}}{\mathrm{3}}{sin}\left({c}\right)+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}{cos}\left({c}\right)\right){cosC}\right) \\ $$$$\mathrm{1}−\mathrm{3}{cos}\left(\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{2}{c}\right)=\mathrm{1}−{cos}\left(\mathrm{2}{c}\right)+\mathrm{3}.\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}{sin}\left(\mathrm{2}{c}\right) \\ $$$$=\mathrm{2}{sin}^{\mathrm{2}} \left({c}\right)−\mathrm{4}\sqrt{\mathrm{2}}{sin}\left({c}\right){cos}\left({c}\right)={sin}\left({c}\right)\left(\mathrm{2}{sin}\left({c}\right)+\mathrm{4}\sqrt{\mathrm{2}}{cos}\left({c}\right)\right) \\ $$$$\frac{\left.\mathrm{1}−\mathrm{3}{cos}\left({B}+{C}\right)\right)}{\mathrm{6}{sin}\left({B}\right){cos}\left({C}\right)}=\frac{{sin}\left({C}\right)\left(\mathrm{2}{sin}\left({C}\right)+\mathrm{4}\sqrt{\mathrm{2}}{cos}\left({C}\right)\right)}{\left(\mathrm{2}{sin}\left({C}\right)+\mathrm{4}\sqrt{\mathrm{2}}{cos}\left({C}\right)\right){cos}\left({C}\right)}=\mathrm{tan}\:\left({C}\right) \\ $$
Commented by MM42 last updated on 15/Jul/24
⋛
$$\:\underline{\underbrace{\lesseqgtr}} \\ $$
Answered by MM42 last updated on 15/Jul/24
((3((1/3)−cos(B+C)))/(6sinBcosC))=((3(cos(B−C)−cos(B+C)))/(6sinBcosC)) =((3[(−2)sin(((B−C−B−C))/2))sin(((B−C+B+C)/2)))/(6sinBcosC)) =((6sinBsinc)/(6sinBcosC))=tanC ✓
$$\frac{\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{3}}−{cos}\left({B}+{C}\right)\right)}{\mathrm{6}{sinBcosC}}=\frac{\mathrm{3}\left({cos}\left({B}−{C}\right)−{cos}\left({B}+{C}\right)\right)}{\mathrm{6}{sinBcosC}} \\ $$$$=\frac{\mathrm{3}\left[\left(−\mathrm{2}\right){sin}\left(\frac{\left.{B}−{C}−{B}−{C}\right)}{\mathrm{2}}\right){sin}\left(\frac{{B}−{C}+{B}+{C}}{\mathrm{2}}\right)\right.}{\mathrm{6}{sinBcosC}} \\ $$$$=\frac{\mathrm{6}{sinBsinc}}{\mathrm{6}{sinBcosC}}={tanC}\:\checkmark \\ $$$$ \\ $$

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