Answer
See the explanation below.
Work Step by Step
Let us consider the right-hand side of the provided expression:
$\sin 2x+\sin 4x+\sin 6x$
Then, rearranging the terms:
$\sin 4x+\left( \sin 2x+\sin 6x \right)$
One of the sum-to-product formulas is $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$. Therefore, $\sin 2x+\sin 6x$ can be written as provided below:
$\sin 2x+\sin 6x=2\sin \frac{2x+6x}{2}\cos \frac{2x-6x}{2}$
Then, the expression can be computed as shown below:
$\begin{align}
& \sin 4x+\left( \sin 2x+\sin 6x \right)=\sin 4x+2\sin \left( \frac{2x+6x}{2} \right)\cos \left( \frac{2x-6x}{2} \right) \\
& =\sin 4x+2\sin \left( \frac{8x}{2} \right)\cos \left( \frac{-4x}{2} \right) \\
& =\sin 4x+2\sin 4x\cos \left( -2x \right)
\end{align}$
By applying the even-odd identity, which is $cos(-x)=\cos x$, the expression can be further evaluated as given below:
$\begin{align}
& \sin 4x+2\sin 4x\cos \left( -2x \right)=\sin 4x+2\sin 4x\cos 2x \\
& =\sin \left( 2\cdot 2x \right)+2\sin 4x\cos 2x
\end{align}$
Also, apply the double angle formula, which is $\sin 2\theta =2\sin \theta \cos \theta $, so, the expression can be further solved as:
$\sin \left( 2\cdot 2x \right)+2\sin 4x\cos 2x=2\sin 2x\cos 2x+2\sin 4x\cos 2x$
By taking out the common terms:
$2\sin 2x\cos 2x+2\sin 4x\cos 2x=2\cos 2x\left( \sin 2x+\sin 4x \right)$
Again, applying the sum-to-product formula, which is $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$, the expression can be calculated as shown below:
$\begin{align}
& 2\cos 2x\left( \sin 2x+\sin 4x \right)=2\cos 2x\left[ 2\sin \left( \frac{2x+4x}{2} \right)\cos \left( \frac{2x-4x}{2} \right) \right] \\
& =2\cos 2x\cdot 2\sin \left( \frac{6x}{2} \right)\cos \left( \frac{-2x}{2} \right) \\
& =2\cos 2x\cdot 2\sin 3x\cos \left( -x \right)
\end{align}$
Then, apply the even-odd identity, which is $cos(-x)=\cos x$:
$\begin{align}
& 2\cos 2x\cdot 2\sin 3x\cos \left( -x \right)=2\cos 2x\cdot 2\sin 3x\cos x \\
& =4\cos 2x\sin 3x\cos x
\end{align}$
Then rearrange the terms:
$4\cos 2x\sin 3x\cos x=4\cos x\cos 2x\sin 3x$
Thus, the right-hand side of the expression is equal to the left-hand side:
$4\cos x\cos 2x\sin 3x=\sin 2x+\sin 4x+\sin 6x$.
