Animated Solution for Mathematics - Trigonometry: $96\cos \frac{\pi }{33}\cos \frac{2\pi }{33}\cos \frac{4\pi }{33}\cos \frac{8\pi }{33}\cos \frac{16\pi }{33}$ is equal to

Imagine you are staring at a long, intimidating product of trigonometric functions:

At first glance, it looks like a nightmare of irrational numbers. But in the world of JEE Advanced, complexity is often just a mask for elegance. Let's peel back that mask.

Look closely at the angles: $\frac{\pi }{33},\frac{2\pi }{33},\frac{4\pi }{33},\frac{8\pi }{33},\frac{16\pi }{33}$. Do you see the rhythm? Each angle is exactly double the previous one.

This is not a coincidence; it is a mathematical signature. Whenever you see a product of cosines where the arguments follow a doubling pattern, your mind should immediately jump to the product identity:

This formula is your most powerful weapon.

Let's define our variables. Our smallest angle is $A=\frac{\pi }{33}$, and we have $n=5$ terms.

Substituting these into our identity, the expression becomes:

Suddenly, the product has collapsed into a single, manageable fraction. We know that $2^5=32$, so our expression simplifies to:

Dividing the coefficients, $96/32$ gives us a clean $3$. We are left with:

Now, we face the final hurdle. The angles in the numerator and denominator are different.

But look at the numerator: $\frac{32\pi }{33}$. This is just $\pi -\frac{\pi }{33}$.

Using the supplementary angle identity $\sin (\pi -\theta )=\sin \theta$, we can rewrite $\sin (\frac{32\pi }{33})$ as $\sin (\frac{\pi }{33})$. The expression becomes:

The sine terms cancel out perfectly, leaving us with the final answer: 3.

A complex, intimidating product reduced to a single integer. This is the elegance of mathematics—finding the simple truth hidden beneath the surface.