證明題(sin.cos)

sin(B+C-A)-sin(C+A-B)+sin(A+B-C)
= sin(180° - 2A) - sin(180° - 2B) + sin(180° - 2C)
= sin2A - sin2B + sin2C
= 2cos(A+B) sin(A-B) + 2sinC cosC
= 2cos(180° - C) sin(A-B) + 2sin(180° - (A+B)) cosC
= -2cosC sin(A-B) + 2cosC sin(A+B)
= 2cosC [sin(A+B) - sin(A-B)]
= 2cosC [sinAcosB + cosAsinB - (sinAcosB - cosAsinB)]
= 2cosC 2cosAsinB
= 4cosAsinBcosC