Question Number 44096 by peter frank last updated on 21/Sep/18
Answered by $@ty@m last updated on 21/Sep/18
$$\left({a}\right)\:\mathrm{cos}\:{A}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$${a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\mathrm{cos}\:{A} \\ $$$$\Rightarrow{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\left({b}−{c}\right)^{\mathrm{2}} +\mathrm{4}{bc}\mathrm{sin}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}} \\ $$
Commented by math1967 last updated on 21/Sep/18
$${a}^{\mathrm{2}} =\left({b}−{c}\right)^{\mathrm{2}} +\mathrm{4}{bcsin}^{\mathrm{2}} \frac{{A}}{\mathrm{2}} \\ $$
Commented by $@ty@m last updated on 21/Sep/18
$${Thanks}\:{for}\:{correction}. \\ $$$${It}'{s}\:{typing}\:{mistake}. \\ $$
Commented by math1967 last updated on 21/Sep/18
$${It}'{s}\:{ok}\:{sir} \\ $$
Answered by $@ty@m last updated on 21/Sep/18
$$\left({b}\right)\:{Let}\:\frac{\mathrm{sin}\:{A}}{{a}}=\frac{\mathrm{sin}\:{B}}{{b}}=\frac{\mathrm{sin}\:{C}}{{c}}={R} \\ $$$${RHS}=\frac{\mathrm{sin}\:\left({A}−{B}\right)}{\mathrm{sin}\:\left({A}+{B}\right)} \\ $$$$=\frac{\mathrm{sin}\:{A}\mathrm{cos}\:{B}−\mathrm{cos}\:{A}\mathrm{sin}\:{B}}{\mathrm{sin}\:{C}} \\ $$$$=\frac{{aR}\mathrm{cos}\:{B}−{bR}\mathrm{cos}\:{A}}{{cR}} \\ $$$$=\frac{{a}\mathrm{cos}\:{B}−{b}\mathrm{cos}\:{A}}{{c}} \\ $$$$=\frac{{a}×\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}}−{b}×\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}}{{c}} \\ $$$$=\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }{\mathrm{2}{c}^{\mathrm{2}} } \\ $$$$=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} } \\ $$$$={LHS} \\ $$