You should convert into a product the sum `sin 3A + sin 3B ` such that:

`sin 3A + sin 3B = 2 sin ((3A+3B)/2)*cos ((3A-3B)/2)`

Since you need to prove that `2 sin ((3A+3B)/2)*cos ((3A-3B)/2) = 0` , hence, either `sin((3A+3B)/2) = 0` , or `cos((3A-3B)/2) = 0` .

You need to convert sinA + sin B into a product such that:

`sinA + sin B = 2 sin ((A+B)/2)*cos ((A-B)/2)`

`cos B - cos A = 2 sin ((A+B)/2)*sin ((A-B)/2)`

The problem provides the information that:

`2 sin ((A+B)/2)*cos ((A-B)/2) = sqrt3*(2 sin ((A+B)/2)*sin ((A-B)/2))`

You need to reduce by `2 sin ((A+B)/2)` such that:

`cos ((A-B)/2) = sqrt3*(sin ((A-B)/2))`

You need to divide by `cos ((A-B)/2)` both sides such that:

`(sin ((A-B)/2))/cos ((A-B)/2) = 1/sqrt3`

`tan ((A-B)/2) = sqrt3/3 =gt ((A-B)/2) = pi/6`

You need to substitute `pi/6` for `((A-B)/2)` in expression `2 sin ((3A+3B)/2)*cos ((3A-3B)/2)` such that:

`2 sin ((3A+3B)/2)*cos ((3A-3B)/2) = 2 sin ((3A+3B)/2)*cos 3*(pi/6)`

`2 sin ((3A+3B)/2)*cos ((3A-3B)/2) = 2 sin ((3A+3B)/2)*cos (pi/2)`

You should remember that `cos (pi/2) = 0` , hence `2 sin ((3A+3B)/2)*cos (pi/2) = 0` .

**Hence, using the condition of the problem, yields that`((A-B)/2) = pi/6` , thus `sin 3A + sin 3B = 2 sin ((3A+3B)/2)*cos ((3A-3B)/2) = 0.` **

Sin A + Sin B = 3^1/2 (Cos B - Cos A)

(Sin A + Sin B)/(Cos B- Cos A) = 3^1/2

2 Sin (A+B)/2 . Cos (A-B)/2 / 2 Sin (A+B)/2 . Sin (A-B)/2 =3^1/2

Cot (A-B)/2 =3^1/2

Tan (A-B)/2 = 1/3^1/2

(A-B)/2 = Pi/6

A-B= Pi/3

A = Pi/3 + B

3A = Pi + 3B

Sin 3A = - Sin (3B)

Sin 3A + Sin 3B = 0 (Hence Proved)