Animated Solution for Mathematics - Trigonometry: Show that $16\cos (\frac{2\pi }{15})\cos (\frac{4\pi }{15})\cos (\frac{8\pi }{15})\cos (\frac{16\pi }{15})=1$.

Imagine you are standing before a massive, intimidating trigonometric expression: $16\cos (\frac{2\pi }{15})\cos (\frac{4\pi }{15})\cos (\frac{8\pi }{15})\cos (\frac{16\pi }{15})$. At first glance, it looks like a chaotic mess of fractions and cosine functions.

In the world of JEE Advanced mathematics, chaos is often just order in disguise. Our goal is to prove this entire expression equals $1$.

The first step in any complex problem is to observe. Look at the angles inside the cosine functions: $\frac{2\pi }{15}$, $\frac{4\pi }{15}$, $\frac{8\pi }{15}$, and $\frac{16\pi }{15}$.

Each angle is exactly double the previous one. If we define our starting angle as $\theta =\frac{2\pi }{15}$, then the sequence of angles is simply $\theta ,2\theta ,4\theta$, and $8\theta$.

This is a geometric progression with a common ratio of $2$. Recognizing this pattern is the moment the problem shifts from impossible to solvable.

Now that we have identified the pattern, we need the right tool. In trigonometry, there is a powerful identity for products of cosines with doubling angles:

Think of this identity as a compression algorithm. It takes a long, unwieldy product and zips it into a compact, elegant ratio of sine functions. Here, we have $n=4$ terms, so we are ready to apply this formula.

Let us apply our tool with $\theta =\frac{2\pi }{15}$ and $n=4$. Substituting these into our identity, the product of the four cosines becomes:

Remember, our original expression was multiplied by $16$. So, the full expression is:

The $16$ in the numerator and the $16$ in the denominator cancel out with absolute precision. This is the moment of mathematical satisfaction—where the complexity vanishes, leaving us with:

We are almost at the finish line. We have $\sin (\frac{32\pi }{15})$ in the numerator. This angle is larger than $2\pi$, which means it has wrapped around the unit circle.

Let us break it down:

Because the sine function is periodic with a period of $2\pi$, we know that $\sin (2\pi +\alpha )=\sin (\alpha )$. Therefore, $\sin (\frac{32\pi }{15})=\sin (\frac{2\pi }{15})$.

Substituting this back into our fraction, we get:

And there it is. The entire expression collapses to $1$. It is a beautiful reminder that even the most daunting problems yield to the right perspective and the right tools.