Animated Solution for Mathematics - Trigonometry: 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

Welcome, future engineer! Today, we are going to dismantle a trigonometric beast. When you first look at the expression $36(4{\cos }^2{9}^{\circ }-1)(4{\cos }^{2}27^{\circ }-1)(4{\cos }^{2}81^{\circ }-1)(4{\cos }^{2}243^{\circ }-1)$, it is natural to feel a bit overwhelmed.

It looks like a chaotic mess of large angles and squared terms. But in the world of JEE Advanced, complexity is often just a mask for hidden symmetry. Let us peel back that mask together.

The first step in any trigonometric problem is to look at the angles. We have $9^{\circ },27^{\circ },81^{\circ },$ and $243^{\circ }$. Do you see the connection?

$81^{\circ }$ is just $90^{\circ }-9^{\circ }$, and $243^{\circ }$ is $270^{\circ }-27^{\circ }$. This is our golden ticket. Using the co-function identities, we can rewrite $\cos 81^{\circ }$ as $\sin 9^{\circ }$ and $\cos 243^{\circ }$ as $-\sin 27^{\circ }$.

Because our expression involves ${\cos }^{2}243^{\circ }$, the negative sign vanishes upon squaring, leaving us with ${\sin }^{2}27^{\circ }$. Now, our expression looks much friendlier:

Now, let us group the terms by their angles. We pair the $9^{\circ }$ terms and the $27^{\circ }$ terms. We are left with two identical structures of the form $(4{\cos }^2\theta -1)(4{\sin }^2\theta -1)$.

Let us expand this general form. Multiplying these brackets gives:

Factoring out the $-4$, we get $16{\sin }^2\theta {\cos }^2\theta -4({\sin }^2\theta +{\cos }^2\theta )+1$. Since ${\sin }^2\theta +{\cos }^2\theta =1$, this simplifies beautifully to:

We are almost there. Look at $16{\sin }^2\theta {\cos }^2\theta$. We know that $2\sin \theta \cos \theta =\sin 2\theta$. Squaring this gives $4{\sin }^2\theta {\cos }^2\theta ={\sin }^{2}2\theta$.

Therefore, $16{\sin }^2\theta {\cos }^2\theta$ is simply $4{\sin }^{2}2\theta$. Our entire expression has now collapsed into:

This is the elegance of trigonometry—taking a massive, intimidating product and reducing it to a simple calculation.

Now, we just need the standard values. We know $\sin 18^{\circ }=\frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ }=\cos 36^{\circ }=\frac{\sqrt{5}+1}{4}$.

Substituting these into our brackets, we get:

Calculating these, the first bracket becomes $\frac{-6-2\sqrt{5}}{4}$ and the second becomes $\frac{-6+2\sqrt{5}}{4}$. Multiplying these together, we use the difference of squares:

Finally, $36\times 1=36$. And there you have it! A complex problem solved with nothing but fundamental identities and a bit of patience. Keep practicing, and soon, you will see these patterns instantly. The final answer is 36.