Animated Solution for Mathematics - Trigonometry: The value of $2\sin (\frac{\pi }{8})\sin (\frac{2\pi }{8})\sin (\frac{3\pi }{8})\sin (\frac{5\pi }{8})\sin (\frac{6\pi }{8})\sin (\frac{7\pi }{8})$ is:

Imagine you are standing before a complex trigonometric expression. It looks like a wall of numbers, a product of six different sine terms:

At first glance, it is easy to feel overwhelmed. But in the world of JEE Advanced, intimidation is often just a mask for a hidden, elegant simplicity. Let us peel back that mask together.

The Power of Symmetry

Whenever you see a sequence of angles in a trigonometric product, stop and plot them on the unit circle. Look at the angles: $\frac{\pi }{8},\frac{2\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{6\pi }{8},\text{and }\frac{7\pi }{8}$. Notice anything? They are perfectly symmetric around $\frac{\pi }{2}$.

Consider the first and last terms: $\frac{\pi }{8}$ and $\frac{7\pi }{8}$. Their sum is $\frac{\pi }{8}+\frac{7\pi }{8}=\pi$. This is not a coincidence.

Using the identity $\sin (\pi -\theta )=\sin \theta$, we realize that $\sin (\frac{7\pi }{8})=\sin (\frac{\pi }{8})$. We can apply this logic to all the pairs: $\frac{2\pi }{8}+\frac{6\pi }{8}=\pi$ and $\frac{3\pi }{8}+\frac{5\pi }{8}=\pi$.

Suddenly, our massive product collapses into something much more manageable:

Collapsing the Expression

Now, look at the middle term: ${\sin }^2(\frac{2\pi }{8})$. We know that $\frac{2\pi }{8}$ is just $\frac{\pi }{4}$, or $45^{\circ }$.

We know that $\sin (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$, so ${\sin }^2(\frac{\pi }{4})=\frac{1}{2}$. Substituting this back into our expression, the $2$ at the front perfectly cancels with the $\frac{1}{2}$, leaving us with a much cleaner product:

The Hidden Identity

We are left with ${\sin }^2(\frac{\pi }{8}){\sin }^2(\frac{3\pi }{8})$. These are not standard angles, but they are complementary!

Since $\frac{\pi }{8}+\frac{3\pi }{8}=\frac{4\pi }{8}=\frac{\pi }{2}$, this means $\sin (\frac{3\pi }{8})=\cos (\frac{\pi }{8})$. Our expression becomes ${\sin }^2(\frac{\pi }{8}){\cos }^2(\frac{\pi }{8})$, which we can rewrite as:

This is where the magic happens. Recall the double angle formula: $\sin (2\theta )=2\sin \theta \cos \theta$. Rearranging this, we get $\sin \theta \cos \theta =\frac{1}{2}\sin (2\theta )$.

Applying this to our term, we get:

We are almost there. Since $\sin (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$, our expression becomes:

Squaring this, we get $\frac{1}{4\cdot 2}=\frac{1}{8}$.

See how we transformed an intimidating product into a simple fraction? This is the essence of JEE mathematics: not brute force, but the art of seeing the structure beneath the surface. The final answer is $\frac{1}{8}$.