QPaperGen

We use the identity:


\[
\cos^2 A - \cos^2 B = (\cos A + \cos B)(\cos A - \cos B)
\]



Let \( A = 2x \), \( B = 6x \). Then:
\begin{align*}
\text{LHS} &= \cos^2(2x) - \cos^2(6x) \\
&= (\cos 2x + \cos 6x)(\cos 2x - \cos 6x)
\end{align*}

Now apply the sum and difference identities:


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




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



So:
\begin{align*}
\cos 2x + \cos 6x &= 2 \cos(4x) \cos(-2x) = 2 \cos(4x) \cos(2x) \\
\cos 2x - \cos 6x &= -2 \sin(4x) \sin(-2x) = -2 \sin(4x)(- \sin(2x)) = 2 \sin(4x) \sin(2x)
\end{align*}

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

Now use:


\[
\sin(8x) = 2 \sin(4x) \cos(4x), \quad \sin(4x) = 2 \sin(2x) \cos(2x)
\]



So:


\[
\text{LHS} = \sin(8x) \cdot \sin(4x) = \text{RHS}
\]





\[
\boxed{\cos^2(2x) - \cos^2(6x) = \sin(4x) \cdot \sin(8x)}
\]