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Question Number 44096 by peter frank last updated on 21/Sep/18

Answered by $@ty@m last updated on 21/Sep/18

(a) cos A=((b^2 +c^2 −a^2 )/(2bc)) a^2 =b^2 +c^2 −2bccos A ⇒a^2 =b^2 +c^2 −2bc(1−2sin^2 (A/2)) ⇒a^2 =(b−c)^2 +4bcsin^2 (A/2)

$$\left({a}\right)\:\mathrm{cos}\:{A}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$${a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\mathrm{cos}\:{A} \\ $$$$\Rightarrow{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\left({b}−{c}\right)^{\mathrm{2}} +\mathrm{4}{bc}\mathrm{sin}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}} \\ $$

Commented by math1967 last updated on 21/Sep/18

a^2 =(b−c)^2 +4bcsin^2 (A/2)

$${a}^{\mathrm{2}} =\left({b}−{c}\right)^{\mathrm{2}} +\mathrm{4}{bcsin}^{\mathrm{2}} \frac{{A}}{\mathrm{2}} \\ $$

Commented by $@ty@m last updated on 21/Sep/18

Thanks for correction. It′s typing mistake.

$${Thanks}\:{for}\:{correction}. \\ $$$${It}'{s}\:{typing}\:{mistake}. \\ $$

Commented by math1967 last updated on 21/Sep/18

It′s ok sir

$${It}'{s}\:{ok}\:{sir} \\ $$

Answered by $@ty@m last updated on 21/Sep/18

(b) Let ((sin A)/a)=((sin B)/b)=((sin C)/c)=R RHS=((sin (A−B))/(sin (A+B))) =((sin Acos B−cos Asin B)/(sin C)) =((aRcos B−bRcos A)/(cR)) =((acos B−bcos A)/c) =((a×((a^2 +c^2 −b^2 )/(2ac))−b×((b^2 +c^2 −a^2 )/(2bc)))/c) =((a^2 +c^2 −b^2 −b^2 −c^2 +a^2 )/(2c^2 )) =((a^2 −b^2 )/c^2 ) =LHS

$$\left({b}\right)\:{Let}\:\frac{\mathrm{sin}\:{A}}{{a}}=\frac{\mathrm{sin}\:{B}}{{b}}=\frac{\mathrm{sin}\:{C}}{{c}}={R} \\ $$$${RHS}=\frac{\mathrm{sin}\:\left({A}−{B}\right)}{\mathrm{sin}\:\left({A}+{B}\right)} \\ $$$$=\frac{\mathrm{sin}\:{A}\mathrm{cos}\:{B}−\mathrm{cos}\:{A}\mathrm{sin}\:{B}}{\mathrm{sin}\:{C}} \\ $$$$=\frac{{aR}\mathrm{cos}\:{B}−{bR}\mathrm{cos}\:{A}}{{cR}} \\ $$$$=\frac{{a}\mathrm{cos}\:{B}−{b}\mathrm{cos}\:{A}}{{c}} \\ $$$$=\frac{{a}×\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}}−{b}×\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}}{{c}} \\ $$$$=\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }{\mathrm{2}{c}^{\mathrm{2}} } \\ $$$$=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} } \\ $$$$={LHS} \\ $$

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