Question

If $A+B+X=\pi$ then prove that ${\sin }^{3}A\cos \left(B-C\right)+{\sin }^{3}B\cos \left(C-A\right)+{\sin }^{3}C\cos \left(A-B\right)=3\sin A\sin B\sin C$

Answer 1

$si{n}^{3}Acos(B-C)+si{n}^{3}Bcos(C-A)+si{n}^{3}Ccos(A-B)$

$=si{n}^{2}AsinAcos(B-C)+si{n}^{2}BsinBcos(C-A)+si{n}^{2}CsinCcos(A-B)$

$=si{n}^{2}Asin(\pi -(B+C\left)\right)cos(B-C)+si{n}^{2}Bsin(\pi -(C+A\left)\right)cos(C-A)+si{n}^{2}Csin(\pi -(A+B\left)\right)cos(A-B)$ (Using $A+B+C=\pi$)

$=si{n}^{2}Asin(B+C)cos(B-C)+si{n}^{2}Bsin(C+A)cos(C-A)+si{n}^{2}Csin(A+B)cos(A-B)$

(Using $sin(\pi -\theta )=sin\theta$)

$=\frac{si{n}^{2}A(sin2B+sin2C)+si{n}^{2}B(sin2A+sin2C)+si{n}^{2}C(sin2A+sin2B)}{2}$

(Using $sin\theta cos\varphi =\frac{sin(\theta +\varphi )+sin(\theta -\varphi )}{2}$)

$=\frac{si{n}^{2}A2sinBcosB+si{n}^{2}A2sinCcosC+si{n}^{2}B2sinAcosA+si{n}^{2}B2sinCcosC+si{n}^{2}C2sinAcosA+si{n}^{2}C2sinBcosB}{2}$

(Using $sin2\theta =2sin\theta cos\theta$)

$=sinAsinB(sinAcosB+sinBcosA)+sinAsinC(sinAcosC+sinCcosA)+sinBsinC(sinBcosC+sinCcosB)$

$=sinAsinBsin(A+B)+sinAsinCsin(A+C)+sinBsinCsin(B+C)$

(Using $sin(\theta +\varphi )=sin\theta cos\varphi +cos\theta sin\varphi$)

$=sinAsinBsin(\pi -C)+sinAsinCsin(\pi -B)+sinBsinCsin(\pi -A)$

$=3sinAsinBsinC$