To show that [tex]\(\cos^2(3\theta) = \frac{1 + \cos(6\theta)}{2}\)[/tex], we start by applying the appropriate trigonometric identity.
For this task, the double-angle formula for cosine is relevant. The double-angle identity for cosine states:
[tex]\[
\cos(2x) = 2\cos^2(x) - 1
\][/tex]
However, in this context, we want an expression for [tex]\(\cos^2(x)\)[/tex] in terms of double angles. By rearranging the formula, we have:
[tex]\[
2\cos^2(x) = 1 + \cos(2x)
\][/tex]
Dividing both sides by 2, we get:
[tex]\[
\cos^2(x) = \frac{1 + \cos(2x)}{2}
\][/tex]
Now substituting [tex]\(x\)[/tex] with [tex]\(3\theta\)[/tex], we will have:
[tex]\[
x = 3\theta \\
\cos^2(3\theta) = \frac{1 + \cos(2 \cdot 3\theta)}{2}
\][/tex]
Simplifying further, we have:
[tex]\[
\cos^2(3\theta) = \frac{1 + \cos(6\theta)}{2}
\][/tex]
Therefore, we have shown through the double-angle formula application that:
[tex]\[
\cos^2(3\theta) = \frac{1 + \cos(6\theta)}{2}
\][/tex]
