In any triangle ABC, prove that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.​

1 thought on “In any triangle ABC, prove that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.​”

1. [tex]\large\underline{\sf{Solution-}}[/tex]

   Consider,

   [tex] \red{\rm :\longmapsto\: \sf \: asin(B – C) + bsin(C – A) + csin(A – B)}[/tex]

   We know,

   Sine Law

   [tex]\bf :\longmapsto\:In \: \triangle \: ABC[/tex]

   [tex]\rm :\longmapsto\:\dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC} = k[/tex]

   Thus,

   [tex] \red{\rm :\longmapsto\:a = sinA \: } \\ \red{\rm :\longmapsto\: b = sinB\: } \\ \red{\rm :\longmapsto\: c = sinC\: }[/tex]

   Now,

   Consider,

   [tex] \green{\rm :\longmapsto\: \bf \: a \: sin(B – C)}[/tex]

   [tex] \rm \:= \: \:k \: sinA \: sin(B – C)[/tex]

   [tex]\red{\bigg \{ \because \:a = k \: sinA \bigg \}}[/tex]

   [tex] \rm \:= \: \:k \: sin\bigg(\pi – (B + C) \bigg) \: sin(B – C)[/tex]

   [tex]\red{\bigg \{ \because \: A + B + C = \pi\bigg \}}[/tex]

   [tex] \rm \:= \: \:k \: sin(B + C) \: sin(B – C)[/tex]

   [tex]\red{\bigg \{ \because \: sin(\pi – x) = sinx\bigg \}}[/tex]

   [tex] \rm \:= \: \:k( {sin}^{2}B – {sin}^{2}C) [/tex]

   [tex]\bf\implies \:a \: sin(B – C) \: = \: \:k( {sin}^{2}B – {sin}^{2}C) – – (1) [/tex]

   Consider,

   [tex] \green{\rm :\longmapsto\: \bf \: b \: sin(C – A)}[/tex]

   [tex] \rm \:= \: \:k \: sinB \: sin(C – A)[/tex]

   [tex] \rm \:= \: \:k \: sin\bigg(\pi – (C + A) \bigg) \: sin(C – A)[/tex]

   [tex] \rm \:= \: \:k \: sin(C + A) \: sin(C – A)[/tex]

   [tex] \rm \:= \: \:k \: ( {sin}^{2}C – {sin}^{2}A) [/tex]

   [tex]\bf\implies \:b \: sin(C – A) \: = \: \:k( {sin}^{2}C – {sin}^{2}A) – – (2) [/tex]

   Consider,

   [tex] \green{\rm :\longmapsto\: \bf \: c \: sin( A – B)}[/tex]

   [tex] \rm \:= \: \:k \: sinC \: sin(A – B)[/tex]

   [tex] \rm \:= \: \:k \: sin\bigg(\pi – (A + B)\bigg)\: sin(A – B)[/tex]

   [tex] \rm \:= \: \:k \: sin(A + B) \: sin(A – B)[/tex]

   [tex] \rm \:= \: \:k \: ( {sin}^{2}A – {sin}^{2}B) [/tex]

   [tex]\bf\implies \:c \: sin(A – B) \: = \: \:k( {sin}^{2}A – {sin}^{2}B) – – (3) [/tex]

   Hence,

   [tex] \red{\rm :\longmapsto\: \sf \: asin(B – C) + bsin(C – A) + csin(A – B)}[/tex]

   [tex] \rm \:=k( {sin}^{2}B -{sin}^{2}C) + k( {sin}^{2}C-{sin}^{2}A)+\:k( {sin}^{2}A – {sin}^{2}B) [/tex]

   [tex] \rm \:=k( {sin}^{2}B -{sin}^{2}C+{sin}^{2}C-{sin}^{2}A+\:{sin}^{2}A – {sin}^{2}B) [/tex]

   [tex] \rm \:= \: \:k \times 0[/tex]

   [tex] \rm \:= \: \:0[/tex]

   [tex]{{\boxed{\bf{Hence, Proved}}}}[/tex]

   Additional Information :-

   Cosine Law :-

   [tex]\green{\boxed{\bf{cosA = \dfrac{ {b}^{2} + {c}^{2} – {a}^{2} }{2bc} }}}[/tex]

   [tex]\green{\boxed{\bf{cosB = \dfrac{ {c}^{2} + {a}^{2} – {b}^{2} }{2ac} }}}[/tex]

   [tex]\green{\boxed{\bf{cosC = \dfrac{ {a}^{2} + {b}^{2} – {c}^{2} }{2ab} }}}[/tex]

   Projection Formula :-

   [tex]\green{\boxed{\bf{a = b \: cosC + c \: cosB}}}[/tex]

   [tex]\green{\boxed{\bf{b = a \: cosC + c \: cosA}}}[/tex]

   [tex]\green{\boxed{\bf{c = a \: cosB + b \: cosA}}}[/tex]

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