Animated Solution for Mathematics - Trigonometry: The maximum value of $(\cos {\alpha }_1)\cdot (\cos {\alpha }_2)\dots (\cos {\alpha }_n)$, under the restrictions $0\le {\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n\le \pi /2$ and $(\cot {\alpha }_1)\cdot (\cot {\alpha }_2)\dots (\cot {\alpha }_n)=1$ is

Unlocking the Product of Cosines

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of trigonometric variables. We are asked to maximize the product $P={\prod }_{i=1}^n\cos {\alpha }_i$ given the constraint ${\prod }_{i=1}^n\cot {\alpha }_i=1$.

It feels intimidating, doesn't it? But in the world of JEE Advanced, intimidation is often just a mask for elegance. Let us peel back that mask.

The Constraint as a Bridge

We start with the constraint: ${\prod }_{i=1}^n\cot {\alpha }_i=1$. The cotangent function is a ratio, a bridge between the horizontal and the vertical.

Let us write it as:

This is the moment of clarity. If we separate the numerator and the denominator, we get:

Cross-multiplying this gives us the beautiful realization that ${\prod }_{i=1}^n\cos {\alpha }_i={\prod }_{i=1}^n\sin {\alpha }_i$. We have just discovered that the product of our cosines is identical to the product of our sines. This symmetry is the heartbeat of the problem.

The Algebraic Leap

We defined our objective as $P={\prod }_{i=1}^n\cos {\alpha }_i$. Because of our discovery in Phase 1, we now know that $P$ is also equal to ${\prod }_{i=1}^n\sin {\alpha }_i$.

This is where many students get stuck. They try to maximize $P$ directly, but they hit a wall. The trick, the 'Aha!' moment, is to look at $P^2$.

Why? Because $P^2=P\cdot P$. By substituting our two different expressions for $P$, we get:

We can combine these into a single product:

The Double Angle Identity

Now, look at the term inside the product: $\sin {\alpha }_i\cos {\alpha }_i$. Does it ring a bell? It is the ghost of the double angle identity!

We know that $\sin 2{\alpha }_i=2\sin {\alpha }_i\cos {\alpha }_i$, which means $\sin {\alpha }_i\cos {\alpha }_i=\frac{1}{2}\sin 2{\alpha }_i$. Substituting this into our equation for $P^2$, we get:

Since we are multiplying this $n$ times, the factor of $\frac{1}{2}$ comes out as $\frac{1}{2^n}$. Thus:

The Final Optimization

We are almost there. We want to maximize $P^2$. In our expression $P^2=\frac{1}{2^n}{\prod }_{i=1}^n\sin 2{\alpha }_i$, the only variable part is the product of the sines.

We know that the maximum value of any sine function is $1$. Therefore, the product ${\prod }_{i=1}^n\sin 2{\alpha }_i$ is at most $1$.

This maximum is achieved when each $\sin 2{\alpha }_i=1$, which implies $2{\alpha }_i=\frac{\pi }{2}$, or ${\alpha }_i=\frac{\pi }{4}$. Substituting this maximum value back into our equation, we get:

Taking the square root of both sides, we find:

And there it is. The maximum value is $\frac{1}{2^{n/2}}$. You see? The complexity vanished, replaced by the simple, undeniable logic of symmetry and identities. Keep this mindset, and no problem will ever be too large to solve.