Question

sin2A+sin2B+sin2C-sin2(A+B+C)=4sin(B+C)sin(C+A)sin(A+B)

Answer 1

Sin2A+sin2B+sin2C-sin2(A+B+C)=4sin(B+C)sin(C+A)sin(A+B)

$\textbf{To prove:}$

$\sin{2A}+\sin{2B}+\sin{2C}-\sin{2(A+B+C)}=4\,\sin(A+B)\,\sin(B+C)\,\sin(A+C)$

$\textbf{Solution:}$

$\sin{2A}+\sin{2B}+\sin{2C}-\sin{2(A+B+C)}$

$\text{Using the identity}$

$\boxed{\bf\,sinC+sinD=2\,sin(\dfrac{C+D}{2})\,cos(\dfrac{C-D}{2})}$

$\boxed{\bf\,sinC-sinD=2\,cos(\dfrac{C+D}{2})\,sin(\dfrac{C-D}{2})}$

$=2\,\sin(A+B)\,\cos(A-B)+2\,\cos(A+B+2C)\,\sin(-(A+B))$

$=2\,\sin(A+B)\,\cos(A-B)-2\,\cos(A+B+2C)\,\sin(A+B)$

$=2\,\sin(A+B)[\cos(A-B)-\cos(A+B+2C)]$

$\text{Using the identity,}$

$\boxed{\bf\,cosC-cosD=-2\,sin(\dfrac{C+D}{2})\,sin(\dfrac{C-D}{2})}$

$=2\,\sin(A+B)[-2\,\sin(A+C)\,\sin(-B-C)]$

$=2\,\sin(A+B)[-2\,\sin(A+C)\,\sin(-(B+C))]$

$=2\,\sin(A+B)[2\,\sin(A+C)\,\sin(B+C)]$

$=4\,\sin(A+B)\,\sin(B+C)\,\sin(A+C)$

$\implies\bf\sin{2A}+\sin{2B}+\sin{2C}-\sin{2{A+B+C)}=4\,\sin(A+B)\,\sin(B+C)\,\sin(A+C)$