Show that $\sin x(\sin 2x+ \sin 4x +\sin 6x) = \sin 3x \sin 4x$ I got a trigonometric question in an exam and it was confused me to solve.The question contains 3 parts.


*

*Show that $\sin 2x + \sin 4x + \sin 6x = (1 + 2\cos 2x) \sin 4x$.

*By using the above show that $\sin x(\sin 2x + \sin 4x + \sin 6x) = \sin 3x \sin 4x$.

*Derive the values for $\sin (\pi/12)$


I solved the 1st part as bellow. But 2nd and 3rd was unable to solve. So I am Looking for a help to solve this. Thanx :)
sin2x+sin4x+sin6x=(1+2cos2x)sin4x
Using sum to product identities
LHS=sin6x+sin2x+sin4x
sinα+sinβ=2 sin⁡((α+β)/2)  cos⁡((α-β)/2)
LHS=sin4x+2 sin⁡((6x+2x)/2)  cos⁡((6x-2x)/2)
=sin4x+(2 sin⁡4xcos2x)
=sin4x(1+2cos2x)


As a direct solution to the second part, you can use $$\sin(a) \sin(b) =\frac 12(\cos(a-b)-\cos(a+b))$$
to get 
$$
\sin x (\sin 2x+ \sin 4x+\sin6x) = \sin x \sin 2x + \sin x \sin 4x + \sin x \sin 6x = \frac 12(\cos x - \cos 3x) + \frac 12(\cos 3x -\cos 5x) 
+\frac 12 (\cos 5x -\cos 7x) = \frac 12(\cos x -\cos 7x) = \sin 3x \sin 4x.
$$
