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)

 A: Once you are able to write down $\sin(3x)$ in the following way, you are good to go.
$$
\begin{align}
\sin(3x) &= \sin(x+2x)\\
& = \sin(x)\cos(2x) + \cos(x)\sin(2x)\\
&=\sin(x) \cos(2x) + \cos(x)2\sin(x)\cos(x)\\
&=\sin(x)  (\cos(2x) + 2\cos^2(x)) \\
&=\sin(x)  (\cos(2x) + 2\cos^2(x) -1 +1) \\
&=\sin(x)  (\cos(2x) + \cos(2x) +1)\\
&=\sin(x) (1 + 2\cos(2x) )
\end{align}
$$
A: 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 
$$
\sin x (\sin 2x+ \sin 4x+\sin6x) = \sin x \sin 2x + \sin x \sin 4x + \sin x \sin 6x = \frac 12(\cos x - \cos 3x) + \frac 12(\cos 3x -\cos 5x) 
+\frac 12 (\cos 5x -\cos 7x) = \frac 12(\cos x -\cos 7x) = \sin 3x \sin 4x.
$$
A: Well, do as it says and use the first part. You already have that
$$\sin(2x)+\sin(4x)+\sin(6x)=\left(1+2\cos(2x)\right)\sin(4x)$$
So for the case of $\sin(x)\left(\sin(2x)+\sin(4x)+\sin(6x)\right)$, just plug in the result you got from part I:
$$
\begin{align}
\sin(x)\left(\sin(2x)+\sin(4x)+\sin(6x)\right)&=\sin(x)\left(1+2\cos(2x)\right)\sin(4x)\\
&=\left(\sin(x)+2\sin(x)\cos(2x)\right)\sin(4x)\\
&=\left(\sin(x)+2\sin(x)\left(1-2\sin(x)^2\right)\right)\sin(4x)\\
&=\left(3\sin(x)-4\sin(x)^3\right)\sin(4x)\\
&=\sin(3x)\sin(4x)
\end{align}$$
Note that the last identity that I used was the triple angle identity:
$$3\sin(x)-4\sin(x)^3=\sin(3x)$$
For the third part, you want $\sin\left(\frac{\pi}{12}\right)$, but you already know $\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}$, $\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$, $\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$, and $\sin\left(\frac{\pi}{2}\right)$=1. So, plug in $\frac{\pi}{12}$ for $x$ in equation II and solve for it, since every other equation will yield a known value.
