Question

$Ina\Delta ABC,\text{prove that}si{n}^{3}Acos(B-C)+si{n}^{3}Bcos(C-A)+si{n}^{3}Ccos(A-B)=3sinAsinBsinC$

Answer 1

$si{n}^{3}Acos(B-C)+si{n}^{3}Bcos(C-A)+si{n}^{3}Ccos(A-B)=si{n}^{2}AsinAcos(B-C)+si{n}^{2}BsinBcos(C-A)+si{n}^2.sinC(A-B)=si{n}^{2}Asin(\pi -(B+C))cos(B-C)+si{n}^{2}B.sin(\pi -(A+C)).cos(C-A)+si{n}^{2}C.sin(\pi -(A+B)).cos(A-B)=si{n}^{2}Asin(B+C)cos(B-C)+si{n}^{2}B.sin(C+A).cos(C-A)+si{n}^{2}C.sin(A+B)cos(A-B)=si{n}^{2}A.(sin2B+sin2C)+si{n}^{2}B.(sin2C+sin2A)+si{n}^{2}C.(sin2A+sin2B)+si{n}^{2}C.(2sinAcosA+2sinBcosB)=si{n}^{2}A.(2sinBcosB+2sinCcosC)+si{n}^{2}B.(2sinCcosC+2sinAcosA)+si{n}^{2}C.(2sinAcosA+2sinBcosB)=si{n}^{2}A(2sinBcosB+2sinCcosC)+si{n}^{2}B(2sinCcosC+2sinAcosA)+si{n}^{2}C(2sinAcosA+2sinBcosB)=si{n}^{2}A.(2sinBcosB+si{n}^{2}A.2sinCcosC)+si{n}^{2}B.2sinCcosC+si{n}^{2}B.2sinAcosAcosA+si{n}^{2}C.2sinAcosA+si{n}^{2}C.2sinBcosB=k^2{a}^{2}2kbcosB+k^2{a}^2.2bccosC+k^2{b}^2.2kacosC+k^2{b}^2.2kacosA+k^2{c}^2.2kacosA+k^2{c}^2.2kbcosB=k^{3}ab(acosB+bcosA)+k^{3}ac(acosC+ccosA)+k^{3}bc(ccosB+bcosC)=k^{3}abc+k^{3}acb+k^{3}bca=k^{3}3abc=3(ksinA.ksinB.ksinC)=3abc=RHS$