Question

In a triangle $ABC,a^{3}cos(B-C)+b^{3}cos(C-A)+c^{3}cos(A-B)=$

• A. abc
• B. 3abc
• C. a+b+c
• D. None of these

Answer 1

The correct option is B

3abc

$a^{3}cos(B-C)+b^{3}cos(C-A)+c^{3}cos(A-B)=k^{3}si{n}^{3}Acos(B-C)+k^{3}si{n}^{3}Bcos(C-A)+k^{3}si{n}^{3}Ccos(A-B)=\frac{1}{2}{k}^3\left[si{n}^{2}A\right(sin2B+sin2C)+si{n}^{2}B(sin2C+sin2A\left)\right]+si{n}^{2}C(sin2A+sin2B)=k^3\left[si{n}^{2}A\right(sinBcosB+sinCcosC)+si{n}^{2}B(sinCcosC+sinAcosA)+si{n}^{2}C(sinAcosA+sinBcosB\left)\right]=k^3\left[sinAsinB\right(sinAcosB+cosAsinB)+sinBsinC(sinBcosC+cosBsinC)+sinCsinA(sinCcosA+cosCsinA)=k^3[sinAsinBsinC+sinAsinBsinC+sinAsinBsinC]=3k^{3}sinAsinBsinC=3abc$.