Question

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

Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\red{\rm :\longmapsto\: \sf \: asin(B - C) + bsin(C - A) + csin(A - B)}$

We know,

Sine Law

$\bf :\longmapsto\:In \: \triangle \: ABC$

$\rm :\longmapsto\:\dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC} = k$

Thus,

$\red{\rm :\longmapsto\:a = sinA \: } \\ \red{\rm :\longmapsto\: b = sinB\: } \\ \red{\rm :\longmapsto\: c = sinC\: }$

Now,

Consider,

$\green{\rm :\longmapsto\: \bf \: a \: sin(B - C)}$

$\rm \:= \: \:k \: sinA \: sin(B - C)$

$\red{\bigg \{ \because \:a = k \: sinA \bigg \}}$

$\rm \:= \: \:k \: sin\bigg(\pi - (B + C) \bigg) \: sin(B - C)$

$\red{\bigg \{ \because \: A + B + C = \pi\bigg \}}$

$\rm \:= \: \:k \: sin(B + C) \: sin(B - C)$

$\red{\bigg \{ \because \: sin(\pi - x) = sinx\bigg \}}$

$\rm \:= \: \:k( {sin}^{2}B - {sin}^{2}C)$

$\bf\implies \:a \: sin(B - C) \: = \: \:k( {sin}^{2}B - {sin}^{2}C) - - (1)$

Consider,

$\green{\rm :\longmapsto\: \bf \: b \: sin(C - A)}$

$\rm \:= \: \:k \: sinB \: sin(C - A)$

$\rm \:= \: \:k \: sin\bigg(\pi - (C + A) \bigg) \: sin(C - A)$

$\rm \:= \: \:k \: sin(C + A) \: sin(C - A)$

$\rm \:= \: \:k \: ( {sin}^{2}C - {sin}^{2}A)$

$\bf\implies \:b \: sin(C - A) \: = \: \:k( {sin}^{2}C - {sin}^{2}A) - - (2)$

Consider,

$\green{\rm :\longmapsto\: \bf \: c \: sin( A - B)}$

$\rm \:= \: \:k \: sinC \: sin(A - B)$

$\rm \:= \: \:k \: sin\bigg(\pi - (A + B)\bigg)\: sin(A - B)$

$\rm \:= \: \:k \: sin(A + B) \: sin(A - B)$

$\rm \:= \: \:k \: ( {sin}^{2}A - {sin}^{2}B)$

$\bf\implies \:c \: sin(A - B) \: = \: \:k( {sin}^{2}A - {sin}^{2}B) - - (3)$

Hence,

$\red{\rm :\longmapsto\: \sf \: asin(B - C) + bsin(C - A) + csin(A - B)}$

$\rm \:=k( {sin}^{2}B -{sin}^{2}C) + k( {sin}^{2}C-{sin}^{2}A)+\:k( {sin}^{2}A - {sin}^{2}B)$

$\rm \:=k( {sin}^{2}B -{sin}^{2}C+{sin}^{2}C-{sin}^{2}A+\:{sin}^{2}A - {sin}^{2}B)$

$\rm \:= \: \:k \times 0$

$\rm \:= \: \:0$

${{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

Cosine Law :-

$\green{\boxed{\bf{cosA = \dfrac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} }}}$

$\green{\boxed{\bf{cosB = \dfrac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ac} }}}$

$\green{\boxed{\bf{cosC = \dfrac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} }}}$

Projection Formula :-

$\green{\boxed{\bf{a = b \: cosC + c \: cosB}}}$

$\green{\boxed{\bf{b = a \: cosC + c \: cosA}}}$

$\green{\boxed{\bf{c = a \: cosB + b \: cosA}}}$