Prove that (b – a cos C) sin A = a cos A sin C.
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Answer:
sin
3
Acos(B−C)=sin
2
AsinAcos(B−C)
=
2
1
sin
2
A⋅2sin(B+C)cos(B−C)
=
2
1
sin
2
A(sin2B+sin2C)
=sin
2
A(sinBcosB+sinCcosC).
∴L.H.S.=∑sin
3
Acos(B−C)
=sin
2
A(sinBcosB+sinCcosC)+sin
2
B(sinCcosC+sinAcosA)+sin
2
C(sinAcosA+sinBcosB)
=sinAsinB(sinAcosB+cosAsinB)+...+...
=sinAsinBsin(A+B)+...+...
=sinAsinBsinC+...+...
=3sinAsinBsinC.
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