Find the value of `(4cos^(2)9^(@)-1)(4cos^(2)27^(@)-1)`
`(4cos^(2)81^(@)-1)(4cos^(2)243^(@)-1)`.

Explanation

To solve the problem, we need to evaluate the expression:

\[ \frac{(4 \cos^2 9^\circ - 1)(4 \cos^2 27^\circ - 1)}{(4 \cos^2 81^\circ - 1)(4 \cos^2 243^\circ - 1)} \]

### Step 1: Use the identity \(4 \cos^2 \theta - 1 = \sin 3\theta / \sin \theta\)

We can rewrite \(4 \cos^2 \theta - 1\) using the identity:
\[ 4 \cos^2 \theta - 1 = 3 - 4 \sin^2 \theta = \frac{\sin 3\theta}{\sin \theta} \]

### Step 2: Substitute the angles into the identity

Now, we substitute the angles into the identity:

1. For \(4 \cos^2 9^\circ - 1\):
\[ 4 \cos^2 9^\circ - 1 = \frac{\sin 27^\circ}{\sin 9^\circ} \]

2. For \(4 \cos^2 27^\circ - 1\):
\[ 4 \cos^2 27^\circ - 1 = \frac{\sin 81^\circ}{\sin 27^\circ} \]

3. For \(4 \cos^2 81^\circ - 1\):
\[ 4 \cos^2 81^\circ - 1 = \frac{\sin 243^\circ}{\sin 81^\circ} \]

4. For \(4 \cos^2 243^\circ - 1\):
\[ 4 \cos^2 243^\circ - 1 = \frac{\sin 729^\circ}{\sin 243^\circ} \]

### Step 3: Substitute these into the original expression

Now substituting these values into the original expression gives us:
\[ \frac{\left(\frac{\sin 27^\circ}{\sin 9^\circ}\right) \left(\frac{\sin 81^\circ}{\sin 27^\circ}\right)}{\left(\frac{\sin 243^\circ}{\sin 81^\circ}\right) \left(\frac{\sin 729^\circ}{\sin 243^\circ}\right)} \]

### Step 4: Simplify the expression

Now we can simplify the expression:
- The \(\sin 27^\circ\) in the numerator and denominator cancels out.
- The \(\sin 81^\circ\) in the numerator and denominator cancels out.
- This simplifies to:
\[ \frac{\sin 729^\circ}{\sin 9^\circ} \]

### Step 5: Simplify \(\sin 729^\circ\)

Using the periodic property of sine:
\[ \sin 729^\circ = \sin(729 - 720)^\circ = \sin 9^\circ \]

### Step 6: Final simplification

Now substituting back, we have:
\[ \frac{\sin 9^\circ}{\sin 9^\circ} = 1 \]

Thus, the value of the original expression is:
\[ \boxed{1} \]