Question

In ΔABC, prove that ∑a³ sin(B - C) = 0.

Answer 1

Answer:

Step-by-step explanation:

Formula used:

In triangle ABC,

(a/SinA)=(b/sinB)=(c/sinC)=2R

${sin}^2A-{sin}^2B=sin(A+B).sin((A-B)$

${a}^3sin(B-C)\\={(2R.sinA)}^3sin(A-B)\\={2R.}^3.{sin}^3A.sin(B-C)\\={(2R)}^3.{sin}^3A.sin(B-C)\\={(2R)}^3.{sin}^2A.sinA.sin(B-C)\\={(2R)}^3.{sin}^2A.sin(180-(B+C)).sin(B-C)\\={(2R)}^3.{sin}^2A.sin(B+C).sin(B-C)\\={(2R)}^3.{sin}^2A.[{sin}^2B-{sin}^2C}]$

${a}^3sin(B-C)={(2R)}^3[{sin}^2A.{sin}^2B- {sin}^2A.{sin}^2C]\\similarly\\{b}^3sin(C-A)={(2R)}^3[{sin}^2B.{sin}^2C- {sin}^2B.{sin}^2A]\\{c}^3sin(A-B)={(2R)}^3[{sin}^2C.{sin}^2A- {sin}^2C.{sin}^2B]$

$Adding\:these\:3\:equations\:we\:get\\{a}^3sin(B-C)+{b}^3sin(C-A)+{c}^3sin(A-B)\\={(2R)}^3[{sin}^2A.{sin}^2B- {sin}^2A{sin}^C+{sin}^2B.{sin}^2C- {sin}^2B{sin}^2A+.{sin}^2C.{sin}^2A- {sin}^2C{sin}^B]\\={(2R)}^3[0]=0$