You may try also the following approach, hence, you may convert the product `sin(pi/3 - a)sin(pi/3 + a)` into a difference of cosines, using the formula, such that:

`cos a - cos b = -2sin((a+b)/2)sin((a-b)/2)`

Reasoning by analogy, yields:

`sin(pi/3 - a)sin(pi/3 + a) = (-1/2)(cos (2pi/3) - cos 2a)`

Evaluating `cos (2pi/3)` yields:

`cos (2pi/3) = 2cos^2(pi/3) - 1`

`cos (2pi/3) = 2*(1/4) - 1`

`cos (2pi/3) = - 1/2`

`sin(pi/3 - a)sin(pi/3 + a) = (-1/2)(-1/2 - cos 2a)`

Replacing ` (-1/2)(-1/2 - cos 2a)` for `sin(pi/3 - a)sin(pi/3 + a)` in the idenitity you need to prove, yields:

`sin a*(-1/2)(-1/2 - cos 2a) = (1/4)sin 3a`

Reducing duplicate factors yields:

`-sin a*(-1/2 - cos 2a) = (1/2)sin 3a`

`sin a*(1/2 + cos 2a) = (1/2)sin 3a`

Replacing, to the left, `1 - 2sin^2 a` for `cos 2a` yields:

`sin a*(1/2 + 1 - 2sin^2 a) = (1/2)sin 3a`

`sin a*(3/2 - 2sin^2 a) = (1/2)sin 3a`

Multiplying 2 both sides, yields:

`2sin a*(3/2 - 2sin^2 a) = sin 3a`

Performing the multiplication to the left side, yields:

`3sin a - 4sin^3 a = sin 3a`

**You need to notice that the last line represents the triple angle identity, hence, the identity `sin a*sin(pi/3 - a)sin(pi/3 + a) = (1/4) sin 3a` , holds.**

**Further Reading**

Let's first remember the formula for sine of triple angle

`sin3a=3cos^2asina-sin^3a` **(1)**

Now we start from the left hand side and try to get right hand side. We first use addition formula for sine.

`sin(xpmy)=sinxcosy pm cosxsiny`

Hence we have

`sinasin(pi/3-a)sin(pi/3+a)=`

`sina(sin(pi/3)cosa-cos(pi/3)sina)(sin(pi/3)cosa+cos(pi/3)sina)=`

Now we use the fact that `sin(pi/3)=sqrt3/2` and `cos(pi/3)=1/2.`

`sina(sqrt3/2cosa-1/2sina)(sqrt3/2cosa+1/2sina)=`

`sina(3/4cos^2a-1/4sin^2a)=`

`1/4(3cos^2asina-sin^3a)=`

Now we apply formula (1).

`1/4sin3a`

Q.E.D.