數學問題(sin)

A/sin(x-a) + B/sin(x-b)
=[Asin(x-b)+Bsin(x-a)]/[sin(x-a)sin(x-b)]
∴Asin(x-b)+Bsin(x-a)=sinx
∴sinx=A(sinxcosb-cosxsinb)+B(sinxcosa-cosxsina)
RHS=sinx(Acosb+Bcosa)-cosx(Asinb+Bsina)
By combining coeff.
Acosb+Bcosa=1---------(1), Asinb+Bsina=0---------(2)
From (1), Acosb=1-Bcosa
From (2), Asinb=-Bsina
∴tanb=Bsina/(1-Bcosa)
tanb(1-Bcosa)=Bsina/tanb
tanb=Bsina/tanb+Bcosatanb
∴B=tanb/(sina/tanb+cosatanb)
B=tanb/(sina+cosatanb)
From (2), A=-Bsina/sinb
A=-tanbsina/[sinb(sina+cosatanb)]

2008-06-22 17:26:50 補充：
It should be
∴tanb=-Bsina/(1-Bcosa)

tanb(1-Bcosa)=-Bsina/tanb

tanb=-Bsina/tanb+Bcosatanb

∴B=tanb/(-sina/tanb+cosatanb)

B=tanb/(cosatanb-sina)

From (2), A=-Bsina/sinb

A=-tanbsina/[sinb(cosatanb-sina)]

=tanbsina/[sinb(sina-cosatanb)]