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Question Number 21431 by Tinkutara last updated on 23/Sep/17
In any ΔABC, a(b cos C − c cos B) =
$$\mathrm{In}\:\mathrm{any}\:\Delta{ABC},\:{a}\left({b}\:\mathrm{cos}\:{C}\:−\:{c}\:\mathrm{cos}\:{B}\right)\:= \\ $$
Answered by $@ty@m last updated on 23/Sep/17
=abcos C−accos B =((a^2 +b^2 −c^2 )/2)−((c^2 +a^2 −b^2 )/2) =((a^2 +b^2 −c^2 −c^2 −a^2 +b^2 )/2) =b^2 −c^2
$$={ab}\mathrm{cos}\:{C}−{ac}\mathrm{cos}\:{B} \\ $$$$=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}}−\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} −{c}^{\mathrm{2}} −{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$={b}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$
Commented by Tinkutara last updated on 23/Sep/17
I also got this but book says 0.
$$\mathrm{I}\:\mathrm{also}\:\mathrm{got}\:\mathrm{this}\:\mathrm{but}\:\mathrm{book}\:\mathrm{says}\:\mathrm{0}. \\ $$
Answered by behi8341.7@gmail.com last updated on 23/Sep/17
a=b.cosC+c.cosB bcosC−c.cosB=2R(sinBcosC−sinC.cosB)= =2Rsin(B−C)=0⇒B=C (only in isoscale triangle) ⇒LHS=2RsinA.2R.sin(B−C) =4R^2 .sin(180−B−C)sin(B−C)= =4R^2 .sin(B+C).sin(B−C)= =4R^2 .((cos(B−C−B−C)−cos(B−C+B+C))/2)= =2R^2 .(cos2C−cos2B)= =2R^2 [2cos^2 C−1−2cos^2 B+1]= =4R^2 (cos^2 C−cos^2 B)≠0(in general)
$$\mathrm{a}=\mathrm{b}.\mathrm{cosC}+\mathrm{c}.\mathrm{cosB} \\ $$$$\mathrm{bcosC}−\mathrm{c}.\mathrm{cosB}=\mathrm{2R}\left(\mathrm{sinBcosC}−\mathrm{sinC}.\mathrm{cosB}\right)= \\ $$$$=\mathrm{2Rsin}\left(\mathrm{B}−\mathrm{C}\right)=\mathrm{0}\Rightarrow\mathrm{B}=\mathrm{C} \\ $$$$\left(\mathrm{only}\:\mathrm{in}\:\mathrm{isoscale}\:\mathrm{triangle}\right) \\ $$$$\Rightarrow\mathrm{LHS}=\mathrm{2RsinA}.\mathrm{2R}.\mathrm{sin}\left(\mathrm{B}−\mathrm{C}\right) \\ $$$$=\mathrm{4R}^{\mathrm{2}} .\mathrm{sin}\left(\mathrm{180}−\mathrm{B}−\mathrm{C}\right)\mathrm{sin}\left(\mathrm{B}−\mathrm{C}\right)= \\ $$$$=\mathrm{4R}^{\mathrm{2}} .\mathrm{sin}\left(\mathrm{B}+\mathrm{C}\right).\mathrm{sin}\left(\mathrm{B}−\mathrm{C}\right)= \\ $$$$=\mathrm{4R}^{\mathrm{2}} .\frac{\mathrm{cos}\left(\mathrm{B}−\mathrm{C}−\mathrm{B}−\mathrm{C}\right)−\mathrm{cos}\left(\mathrm{B}−\mathrm{C}+\mathrm{B}+\mathrm{C}\right)}{\mathrm{2}}= \\ $$$$=\mathrm{2R}^{\mathrm{2}} .\left(\mathrm{cos2C}−\mathrm{cos2B}\right)= \\ $$$$=\mathrm{2R}^{\mathrm{2}} \left[\mathrm{2cos}^{\mathrm{2}} \mathrm{C}−\mathrm{1}−\mathrm{2cos}^{\mathrm{2}} \mathrm{B}+\mathrm{1}\right]= \\ $$$$=\mathrm{4R}^{\mathrm{2}} \left(\mathrm{cos}^{\mathrm{2}} \mathrm{C}−\mathrm{cos}^{\mathrm{2}} \mathrm{B}\right)\neq\mathrm{0}\left(\mathrm{in}\:\mathrm{general}\right) \\ $$
Commented by Tinkutara last updated on 24/Sep/17
Thanks, book has a typo.
$$\mathrm{Thanks},\:\mathrm{book}\:\mathrm{has}\:\mathrm{a}\:\mathrm{typo}. \\ $$
Answered by myintkhaing last updated on 24/Sep/17
=(((a^2 +b^2 −c^2 ))/2) −(((c^2 +a^2 −b^2 ))/2) =((a^2 +b^2 −c^2 −c^2 −a^2 +b^2 )/2) =b^2 −c^2
$$=\frac{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} \right)}{\mathrm{2}}\:−\frac{\left(\mathrm{c}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{2}}\:=\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} \\ $$

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