trigo

LHS
=cos(A+B)cosC - cosAcos(B+C)
=cos(A+B)cosC-sin(A+B)sinC+sin(A+B)sinC
-cosAcos(B+C)+sinAsin(B+C)-sinAsin(B+C)
=cos(A+B+C)+sin(A+B)sinC-cos(A+B+C)-sinAsin(B+C)
=sin(A+B)sinC-sinAsin(B+C)
=RHS

By putting A=0, B=pi/6 and C=pi/3, we can see that
cosAcos(B+C)-sinAsin(B+C) <> sin(A+B)sinC - sinAsin(B+C) .

L.H.S.=cos(A+B)cosC-cosAcos(B+C)
=[cos(A+B-C)+cos(A+B+C)]/2
-[cos(A-B-C)+cos(A+B+C)]/2
=[cos(A+B-C)-cos(A-B-C)]/2
R.H.S.=sin(A+B)sinC-sinAsin(B+C)
=[cos(A+B-C)-cos(A+B+C)]/2
-[cos(A-B-C)-cos(A+B+C)]/2
=[cos(A+B-C)-cos(A-B-C)]/2
∵L.H.S.=R.H.S.
∴cos(A+B)cosC-cosAcos(B+C)=
sin(A+B)sinC-sinAsin(B+C)

cos(A+B)cosC - cosAcos(B+C)

= [cosAcosB - sinAsinB]cosC - cosA[cosBcosC - sinBsinC]

= [cosAcosB - sinAsinB]cosC - cosA[cosBcosC - sinBsinC]

= cosAcosBcosC - sinAsinBcosC - cosAcosBcosC + cosAsinBsinC

= cosAsinBsinC - sinAsinBcosC

sin(A+B)sinC - sinAsin(B+C)

= [sinAcosB + cosAsinB]sinC - sinA[sinBcosC + cosBsinC]

= sinAcosBsinC + cosAsinBsinC - sinAsinBcosC - sinAcosBsinC

= cosAsinBsinC - sinAsinBcosC

So, cos(A+B)cosC - cosAcos(B+C) = sin(A+B)sinC - sinAsin(B+C)