Question Number 21429 by Tinkutara last updated on 23/Sep/17
$$\mathrm{In}\:\mathrm{any}\:\Delta{ABC},\:\Sigma{a}^{\mathrm{2}} \left(\mathrm{sin}\:{B}\:−\:\mathrm{sin}\:{C}\right)\:= \\ $$
Answered by $@ty@m last updated on 23/Sep/17
![We have (a/(sin A))=(b/(sin B))=(c/(sin C))=k, say ∴Σa^2 (sin B − sin C) = Σa^2 (((b−c)/k)) =(1/k)Σa^2 (b−c) =(1/k){a^2 (b−c)+b^2 (c−a)+c^2 (a−b)} =(1/k){a^2 (b−c)+b^2 (c−b+b−a)+c^2 (a−b)} =(1/k)[{a^2 (b−c)−b^2 (b−c)}+{−b^2 (a−b)+c^2 (a−b)}] =(1/k){(a^2 −b^2 )(b−c)−(b^2 −c^2 )(a−b)} =(1/k)(a−b)(b−c){a+b−(b+c)} =(1/k)(a−b)(b−c){a−c} =−(sin A−sin B)(b−c)(c−a)](Q21432.png)
$${We}\:{have} \\ $$$$\frac{{a}}{\mathrm{sin}\:{A}}=\frac{{b}}{\mathrm{sin}\:{B}}=\frac{{c}}{\mathrm{sin}\:{C}}={k},\:{say} \\ $$$$\therefore\Sigma{a}^{\mathrm{2}} \left(\mathrm{sin}\:{B}\:−\:\mathrm{sin}\:{C}\right)\:= \\ $$$$\Sigma{a}^{\mathrm{2}} \left(\frac{{b}−{c}}{{k}}\right) \\ $$$$=\frac{\mathrm{1}}{{k}}\Sigma{a}^{\mathrm{2}} \left({b}−{c}\right) \\ $$$$=\frac{\mathrm{1}}{{k}}\left\{{a}^{\mathrm{2}} \left({b}−{c}\right)+{b}^{\mathrm{2}} \left({c}−{a}\right)+{c}^{\mathrm{2}} \left({a}−{b}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{{k}}\left\{{a}^{\mathrm{2}} \left({b}−{c}\right)+{b}^{\mathrm{2}} \left({c}−{b}+{b}−{a}\right)+{c}^{\mathrm{2}} \left({a}−{b}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{{k}}\left[\left\{{a}^{\mathrm{2}} \left({b}−{c}\right)−{b}^{\mathrm{2}} \left({b}−{c}\right)\right\}+\left\{−{b}^{\mathrm{2}} \left({a}−{b}\right)+{c}^{\mathrm{2}} \left({a}−{b}\right)\right\}\right] \\ $$$$=\frac{\mathrm{1}}{{k}}\left\{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({b}−{c}\right)−\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\left({a}−{b}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{{k}}\left({a}−{b}\right)\left({b}−{c}\right)\left\{{a}+{b}−\left({b}+{c}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{{k}}\left({a}−{b}\right)\left({b}−{c}\right)\left\{{a}−{c}\right\} \\ $$$$=−\left(\mathrm{sin}\:{A}−\mathrm{sin}\:{B}\right)\left({b}−{c}\right)\left({c}−{a}\right) \\ $$
Commented by Tinkutara last updated on 23/Sep/17
$$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{a}\:\mathrm{whole}\:\mathrm{number},\:\mathrm{not} \\ $$$$\mathrm{this}\:\mathrm{expression}.\:\mathrm{Maybe}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{simplified}\:\mathrm{further}. \\ $$