Can you verify this identity?

cos(4x)cos(2x)= (1/2)[cos8x+cos2x]

To verify this identity, we can make use of the trigonometric identity:

cos(a + b) = cos(a)cos(b) - sin(a)sin(b).

Let's apply this identity to the right side of the equation:

\[ \begin{aligned}
\frac{1}{2}[\cos(8x) + \cos(2x)] &= \frac{1}{2}[\cos(4x + 4x) + \cos(4x - 2x)] \\
&= \frac{1}{2}[\cos(4x)\cos(4x) - \sin(4x)\sin(4x) + \cos(4x)\cos(2x) - \sin(4x)\sin(2x)].
\end{aligned} \]

Now, we can observe that \(\cos(4x)\cos(4x) - \sin(4x)\sin(4x) = 1\), as it is a well-known trigonometric identity.

Simplifying further:

\[ \begin{aligned}
\frac{1}{2}[\cos(4x)\cos(4x) - \sin(4x)\sin(4x) + \cos(4x)\cos(2x) - \sin(4x)\sin(2x)] \\
= \frac{1}{2}[1 + \cos(4x)\cos(2x) - \sin(4x)\sin(2x)].
\end{aligned} \]

Now, notice that \(\cos(4x)\cos(2x) - \sin(4x)\sin(2x)\) is equivalent to \(\cos(4x + 2x) = \cos(6x)\).

Substituting this back into the equation:

\[ \begin{aligned}
\frac{1}{2}[1 + \cos(6x)] = \frac{1}{2}[\cos(6x) + 1].
\end{aligned} \]

This confirms that the right side of the equation simplifies to \(\frac{1}{2}[\cos(6x) + 1]\), which is equal to the left side of the equation \(\cos(4x)\cos(2x)\). Therefore, the identity is verified.