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)


Once you are able to write down $\sin(3x)$ in the following way, you are good to go.
$$
\begin{align}
\sin(3x) &= \sin(x+2x)\\
& = \sin(x)\cos(2x) + \cos(x)\sin(2x)\\
&=\sin(x) \cos(2x) + \cos(x)2\sin(x)\cos(x)\\
&=\sin(x)  (\cos(2x) + 2\cos^2(x)) \\
&=\sin(x)  (\cos(2x) + 2\cos^2(x) -1 +1) \\
&=\sin(x)  (\cos(2x) + \cos(2x) +1)\\
&=\sin(x) (1 + 2\cos(2x) )
\end{align}
$$
