InterviewSolution
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InterviewSolution
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In any triangle `A B C ,`prove that:`a^3cos(B-C)+b^3cos(C-A)+c^3cos(A-B)=3a b c` |
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Answer» From sine law, we have, `a/sinA = b/sinB = c/sinC = k` `=> a = ksinA, b = ksin B, c = ksinC` Now, `L.H.S. = a^3cos(B-C)+b^3os(C-A)+c^3cos(A-B)` `=a^2(ksinAcos(B-C))+b^2(ksinBcos(C-A))+c^2(ksinCcos(A-B))` `=a^2(ksin(pi-(B+C))cos(B-C))+b^2(ksin(pi-(C+A))cos(C-A))+c^2(ksin(pi-(A+B))cos(A-B))` `=k/2[a^2(2sin(B+C)cos(B-C))+b^2(2sin(C+A)sin(C-A))+c^2(2sin(A+B)cos(A-B))]` `=k/2[a^2(sin2B+sin2C)+b^2(sin2C+sin2A)+c^2(sin2A+sin2B]` `=k/2[a^2(2sinBcosB+2sinCcosC)+b^2(2sinCcosC+2sinAcosA)+c^2(2sinAcosA+2sinBcosB]` `=[a^2(ksinBcosB+ksinCcosC)+b^2(ksinCcosC+ksinAcosA)+c^2(ksinAcosA+ksinBcosB]` `=[a^2(bcosB+c cosC)+b^2( c cosC+acosA)+c^2(acosA+bcosB]` `=[a^2bcosB+a^2c cosC+ b^2c cosC+b^2a cosA+c^2a cosA+c^2bcosB]` `=ab(acosB+bcosA)+bc(bcosC+c cosB) + ac(acosC + c cosA)` Using projection formulas, `=ab(c)+bc(a)+ac(b)` `=3abc = R.H.S.` |
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