Animated Solution for Mathematics - Trigonometry: Prove that the values of the function $\frac{\sin x\cos 3x}{\sin 3x\cos x}$ do not lie between $1/3$ and 3 for any real $x$.

Welcome, fellow traveler on the path of JEE mastery! Today, we are going to dismantle a trigonometric beast. It looks intimidating, but beneath the surface, it is a beautiful exercise in algebraic manipulation and logical deduction.

Our goal is to prove that the function $f(x)=\frac{\sin x\cos 3x}{\sin 3x\cos x}$ never takes any value in the interval $(\frac{1}{3},3)$. Imagine a graph where a horizontal band is completely empty—that is what we are about to prove.

First, let us look at our function: $f(x)=\frac{\sin x\cos 3x}{\sin 3x\cos x}$. Instead of panicking at the sight of mixed angles, let us regroup. We see $\sin x$ and $\cos x$ together, and $\cos 3x$ and $\sin 3x$ together.

We can rewrite this as:

Suddenly, the fog clears! We know that $\frac{\sin x}{\cos x}=\tan x$, and the second term is simply the reciprocal of $\tan 3x$. So, our function simplifies to:

This is the first victory. We have tamed the trigonometric mess into a simple ratio.

Now, we need to express everything in terms of $x$. We recall the triple angle identity for tangent, a vital tool in your JEE arsenal:

Substituting this into our expression for $f(x)$, we get:

Don't be intimidated by the fraction within a fraction. When we flip the denominator, we get:

We can factor $\tan x$ out of the denominator: $3\tan x-{\tan }^{3}x=\tan x(3-{\tan }^{2}x)$. Canceling the $\tan x$ from the numerator and denominator, we are left with a clean, elegant expression:

To make this even easier, let us introduce a dummy variable. Let $u={\tan }^{2}x$. Since $x$ is a real number, ${\tan }^{2}x$ must be greater than or equal to zero. Thus, $u\ge 0$.

Our function now looks like this:

Our objective is to find the range of $y$. To do this, we express $u$ in terms of $y$ by cross-multiplying:

We know that $u\ge 0$. Therefore, our expression for $u$ must also be greater than or equal to zero:

To solve this inequality, we multiply by $-1$ to make the coefficients of $y$ positive, which flips the inequality sign:

Using the wavy curve method, we find the critical points at $y=\frac{1}{3}$ and $y=3$. The inequality holds when $y\in (-\infty ,\frac{1}{3}]\cup (3,\infty )$.

This means that $y$ can be any value except those strictly between $\frac{1}{3}$ and $3$. We have successfully identified the 'forbidden zone' and proved that our function avoids it entirely. You have just navigated a complex trigonometric proof with algebraic precision.