QPaperGen

We begin by grouping terms:
\begin{align*}
\text{LHS} &= \sin 2x + \sin 6x + 2\sin 4x
\end{align*}

Apply the identity:


\[
\sin A + \sin B = 2 \sin\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right)
\]



Let \( A = 2x \), \( B = 6x \):
\begin{align*}
\sin 2x + \sin 6x &= 2 \sin\left( \frac{2x + 6x}{2} \right) \cos\left( \frac{2x - 6x}{2} \right) \\
&= 2 \sin(4x) \cos(-2x) = 2 \sin(4x) \cos(2x)
\end{align*}

So:
\begin{align*}
\text{LHS} &= 2 \sin(4x) \cos(2x) + 2 \sin(4x) \\
&= 2 \sin(4x) \left( \cos(2x) + 1 \right)
\end{align*}

Now use the identity:


\[
\cos(2x) + 1 = 2 \cos^2 x
\]



Therefore:
\begin{align*}
\text{LHS} &= 2 \sin(4x) \cdot 2 \cos^2 x \\
&= 4 \cos^2 x \cdot \sin(4x) = \text{RHS}
\end{align*}



\[
\boxed{\sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \cdot \sin 4x} \qquad
\textbf{Hence proved.}
\]