Question Number 20019 by Tinkutara last updated on 20/Aug/17
$$\mathrm{In}\:\mathrm{any}\:\mathrm{triangle}\:{ABC},\:\mathrm{prove}\:\mathrm{that} \\ $$$${a}\:\left(\mathrm{cos}\:{C}\:−\:\mathrm{cos}\:{B}\right)\:=\:\mathrm{2}\:\left({b}\:−\:{c}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}} \\ $$
Answered by ajfour last updated on 20/Aug/17
![l.h.s.=a[((a^2 +b^2 −c^2 )/(2ab))−((c^2 +a^2 −b^2 )/(2ac))] =((c[a^2 +(b^2 −c^2 )]+b[(b^2 −c^2 )−a^2 ])/(2bc)) =(((b^2 −c^2 )(b+c)−a^2 (b−c))/(2bc)) =(((b−c)[(b+c)^2 −a^2 ])/(2bc)) =(b−c)(((2bc+b^2 +c^2 −a^2 ))/(2bc)) =(b−c)(1+((b^2 +c^2 −a^2 )/(2bc))) =(b−c)(1+cos A) =2(b−c)cos^2 (A/2) = r.h.s.](https://www.tinkutara.com/question/Q20023.png)
$${l}.{h}.{s}.={a}\left[\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}−\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}}\right] \\ $$$$\:\:=\frac{{c}\left[{a}^{\mathrm{2}} +\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\right]+{b}\left[\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} \right]}{\mathrm{2}{bc}} \\ $$$$\:\:=\frac{\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\left({b}+{c}\right)−{a}^{\mathrm{2}} \left({b}−{c}\right)}{\mathrm{2}{bc}} \\ $$$$\:\:=\frac{\left({b}−{c}\right)\left[\left({b}+{c}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right]}{\mathrm{2}{bc}} \\ $$$$\:\:=\left({b}−{c}\right)\frac{\left(\mathrm{2}{bc}+{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{\mathrm{2}{bc}} \\ $$$$\:\:=\left({b}−{c}\right)\left(\mathrm{1}+\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\right) \\ $$$$\:\:=\left({b}−{c}\right)\left(\mathrm{1}+\mathrm{cos}\:{A}\right) \\ $$$$\:\:=\mathrm{2}\left({b}−{c}\right)\mathrm{cos}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}}\:=\:{r}.{h}.{s}. \\ $$
Commented by Tinkutara last updated on 20/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$