The value of $$36(4\cos^2 9^\circ - 1)(4\cos^2 27^\circ - 1)(4\cos^2 81^\circ - 1)(4\cos^2 243^\circ - 1)$$ is

We use the identity: $$\frac{\sin 3\theta}{\sin\theta} = 3 - 4\sin^2\theta = 4\cos^2\theta - 1$$

Applying to each factor:

$$4\cos^2 9° - 1 = \frac{\sin 27°}{\sin 9°}$$

$$4\cos^2 27° - 1 = \frac{\sin 81°}{\sin 27°}$$

$$4\cos^2 81° - 1 = \frac{\sin 243°}{\sin 81°}$$

$$4\cos^2 243° - 1 = \frac{\sin 729°}{\sin 243°}$$

The product telescopes:

$$\frac{\sin 27°}{\sin 9°} \cdot \frac{\sin 81°}{\sin 27°} \cdot \frac{\sin 243°}{\sin 81°} \cdot \frac{\sin 729°}{\sin 243°} = \frac{\sin 729°}{\sin 9°}$$

Since $$729° = 2 \times 360° + 9°$$, we have $$\sin 729° = \sin 9°$$.

Therefore the product = $$\frac{\sin 9°}{\sin 9°} = 1$$

So $$36 \times 1 = 36$$, which is Option D.