# Question-68210

Question Number 68210 by peter frank last updated on 07/Sep/19
Answered by $@ty@m123 last updated on 08/Sep/19 $${Let}\:\frac{\mathrm{sin}\:{A}}{{a}}=\frac{\mathrm{sin}\:{B}}{{b}}=\frac{\mathrm{sin}\:{C}}{{c}}={R} \\$$$$\Rightarrow\mathrm{sin}\:{A}={aR},\:\mathrm{sin}\:{B}={bR},\:\mathrm{sin}\:{C}={cR}\:…\left(\mathrm{1}\right) \\$$$$\left(\mathrm{2}\right)\:\frac{\mathrm{sin}\:\left({A}−{B}\right)\mathrm{sin}\:{C}}{\mathrm{1}+\mathrm{cos}\:\left({A}−{B}\right)\mathrm{cos}\:{C}} \\$$$$=\:\frac{\mathrm{sin}\:\left({A}−{B}\right)\mathrm{sin}\:\left({A}+{B}\right)}{\mathrm{1}−\mathrm{cos}\:\left({A}−{B}\right)\mathrm{cos}\:\left({A}+{B}\right)} \\$$$$=\frac{\mathrm{sin}\:^{\mathrm{2}} {A}−\mathrm{sin}\:^{\mathrm{2}} {B}}{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {A}+\mathrm{sin}\:^{\mathrm{2}} {B}} \\$$$$=\frac{\mathrm{sin}\:^{\mathrm{2}} {A}−\mathrm{sin}\:^{\mathrm{2}} {B}}{\mathrm{sin}\:\:^{\mathrm{2}} {A}+\mathrm{sin}\:^{\mathrm{2}} {B}} \\$$$$=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\:,\:{using}\:\left(\mathrm{1}\right) \\$$ Commented by peter frank last updated on 08/Sep/19 $${thanks}\:{for}\:{solvinv}\:{sir} \\$$ Commented by$@ty@m123 last updated on 09/Sep/19
$${You}'{re}\:{welcome}. \\$$$${Meanwhile}\:{trying}\:{for}\:\left(\mathrm{1}\right) \\$$
Answered by \$@ty@m123 last updated on 10/Sep/19
$$\left(\mathrm{1}\right)\:\left({a}+{c}\right)\mathrm{cos}\:\frac{{B}}{\mathrm{4}}−\mathrm{2}{c}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}} \\$$$$={R}\left(\mathrm{sin}\:{A}+\mathrm{sin}\:{C}\right)\mathrm{cos}\:\frac{{B}}{\mathrm{4}}−\mathrm{2}{R}\mathrm{sin}\:{C}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}} \\$$$$={R}.\mathrm{2sin}\:\frac{{A}+{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}−\mathrm{2}{R}\mathrm{sin}\:{C}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}} \\$$$$={R}.\mathrm{2sin}\:\frac{\pi−{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}−\mathrm{2}{R}\mathrm{sin}\:{C}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}} \\$$$$={R}.\mathrm{2cos}\:\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}−\mathrm{2}{R}\mathrm{sin}\:{C}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}} \\$$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}\left\{\mathrm{cos}\:\frac{{A}−{C}}{\mathrm{2}}−\mathrm{sin}\:{C}\right\} \\$$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}\left\{\mathrm{cos}\:\frac{{A}−{C}}{\mathrm{2}}−\mathrm{cos}\:\frac{\pi−\mathrm{2}{C}}{\mathrm{2}}\right\} \\$$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{2sin}\:\frac{{A}−{C}+\pi−\mathrm{2}{C}}{\mathrm{2}}\mathrm{sin}\:\frac{\pi−\mathrm{2}{C}−{A}+{C}}{\mathrm{2}} \\$$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{2sin}\:\left(\frac{\pi}{\mathrm{2}}−\frac{{A}−\mathrm{3}{C}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\frac{{A}+{C}}{\mathrm{2}}\right) \\$$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{2cos}\:\left(\frac{{A}−\mathrm{3}{C}}{\mathrm{2}}\right)\mathrm{cos}\:\:\left(\frac{{A}+{C}}{\mathrm{2}}\right) \\$$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{2cos}\:\left(\frac{{A}−\mathrm{3}{C}}{\mathrm{2}}\right)\mathrm{cos}\:\:\left(\frac{\pi−{B}}{\mathrm{2}}\right) \\$$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{2cos}\:\left(\frac{{A}−\mathrm{3}{C}}{\mathrm{2}}\right)\mathrm{sin}\:\:\frac{{B}}{\mathrm{2}} \\$$$$={R}\mathrm{sin}\:{B}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{2cos}\:\left(\frac{{A}−\mathrm{3}{C}}{\mathrm{2}}\right) \\$$$$={b}.\mathrm{2cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{cos}\:\left(\frac{{A}−\mathrm{3}{C}}{\mathrm{2}}\right) \\$$$$={b}.\mathrm{2cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{cos}\:\left(\frac{\pi−\left({B}+{C}\right)−\mathrm{3}{C}}{\mathrm{2}}\right) \\$$$$={b}.\mathrm{2cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{cos}\:\left(\frac{\pi−{B}−\mathrm{4}{C}}{\mathrm{2}}\right) \\$$$$={b}.\mathrm{2cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{sin}\:\:\left(\frac{{B}+\mathrm{4}{C}}{\mathrm{2}}\right) \\$$$${Got}\:{stuck}\:{here}. \\$$$${Either}\:{question}\:{or}\:{my}\:{path}\:{is}\:{wrong}. \\$$
Commented by peter frank last updated on 14/Sep/19
$${thank}\:{you} \\$$$$\\$$
Commented by peter frank last updated on 14/Sep/19
$${thank}\:{you} \\$$$$\\$$