Giải thích các bước giải:

Ta có:

$\sin(A+B-2C)=\sin(A+B+C-3C)=\sin(\pi-3C)=\sin(3C)$

Ta có:

$\cos\dfrac{B+C}2=\cos\dfrac{180^o-A}2=\cos(90^o-\dfrac{A}2)=\sin\dfrac{A}2$

Ta có:

$\sin(A+B)=\sin(180^o-C)=\sin C$

Ta có:

$\cos\dfrac{A+B+2C}2=\cos\dfrac{180^o+C}2=\cos(90^o+\dfrac{C}2)=\sin(90^o-(90^o+\dfrac{C}2))=\sin(-\dfrac{C}2)=-\sin\dfrac{C}2$