First, since the equation is an identity, it has to be true for all values of A, B and C, so let A=B=C, then we have:
sin(3A)+sin(3A)-sin(3A)=4sin(0)cos(0)cos(0),
which reduces to sin(3A)=0 which is not an identity but has a particular solution of A=n(pi)/3, where n is an integer. Therefore, the relationship is generally false and the question has been misstated.
All we can do is to find out what the identity should have been using known trig identities. I therefore offer the following as a restatement of the original problem. I suspect the left-hand side is wrong because of lack of symmetry, while the right-hand side is symmetrical.
Let's look at the sine of the sum of 3 angles and combinations of sum and differences:
(1) sin(X+Y+Z) = sin(X+(Y+Z)) = sinXcos(Y+Z) + cosXsin(Y+Z) = sinXcosYcosZ - sinXsinYsinZ + cosXsinYcosZ + cosXcosYsinX. If we keep X, Y and Z in order we can use c and s in the right order to reduce the notation. So we have sin(X+Y+Z)=scc-sss+ccs-csc+ccs.
(2) sin(X-(Y+Z))=cosXsin(Y+Z)-sinXcos(Y+Z)=csc+ccs-scc+sss.
If we subtract (2) from (1), some terms cancel and others double up: (3) 2scc-2sss.
(4) sin(X+(Y-Z))=sinXcos(Y-Z)+cosXsin(Y-Z)=scc+sss+ccs-csc.
(5) sin(X-(Y-Z))=cosXsin(Y-Z)-sinXcos(Y-Z)=ccs-csc-scc-sss.
Subtract (5) from (4): (6) 2scc+2sss.
Add (3) and (6): (7) 4scc=4sinXcosYcosZ.
Put X=(A-B)/2, Y=(B-C)/2, Z=(C-A)/2 to match the right-hand side of the equation. From these we can work out combinations of X, Y and Z.
X+Y+Z=0 so sin(X+Y+Z)=0;
X-(Y+Z)=(A-B-B+C-C+A)/2=A-B;
X+Y-Z=(A-B+B-C-C-A)/2=A-C;
X-(Y-Z)=(A-B-B+C+C-A)/2=C-B.
Therefore:
4sin((A-B)/2).cos((B-C)/2).cos((C-A)/2)=(3)+(6)=(1)-(2)+(4)-(5)=0-sin(A-B)+sin(A-C)-sin(C-B)=
sin(B-A)+sin(A-C)+sin(B-C) or sin(B-A)+sin(B-C)-sin(C-A).