Animated Solution for Mathematics - Straight Lines: Let the range of the function $f(x)=6+16\cos x\cdot \cos (\frac{\pi }{3}-x)\cdot \cos (\frac{\pi }{3}+x)\cdot \sin 3x\cdot \cos 6x,$ $x\in R$ be $[\alpha ,\beta ]$. Then the distance of the point $(\alpha ,\beta )$ from the line $3x+4y+12=0$ is:

Welcome, future engineer. Today, we face a trigonometric leviathan. At first glance, the function

looks like a nightmare of complexity. But in the world of JEE Advanced, complexity is often just a mask for elegance. Our mission is to strip away this mask.

The secret weapon here is the triple product identity. Look at the first three terms: $\cos x\cdot \cos (\frac{\pi }{3}-x)\cdot \cos (\frac{\pi }{3}+x)$.

This is a classic pattern. We know that

By applying this, the product collapses into $\frac{1}{4}\cos 3x$. Suddenly, the monster is shrinking.

Substituting this back, our function becomes

which simplifies beautifully to

Now, we see a familiar rhythm. We have $\sin 3x$ and $\cos 3x$ together, which screams the double angle formula: $2\sin A\cos A=\sin 2A$.

Let us split the coefficient $4$ into $2\cdot 2$. The expression becomes

The term in the parenthesis is exactly $\sin 6x$. So, we have

But wait, we can do it again! Since $2\sin 6x\cos 6x$ is just $\sin (2\cdot 6x)$, the entire expression has collapsed into

Finding the range is now trivial. The function $\sin 12x$ oscillates between $-1$ and $1$.

Thus, $f(x)$ oscillates between $6-1=5$ and $6+1=7$. Our range $[\alpha ,\beta ]$ is $[5,7]$.

We have our point $(5,7)$. The final step is to find the distance of this point from the line $3x+4y+12=0$.

Using the perpendicular distance formula

we substitute $x_1=5$, $y_1=7$, $a=3$, $b=4$, and $c=12$.

The numerator is

The denominator is $\sqrt{3^2+4^2}=5$. The distance is

We have conquered the leviathan. Remember, in physics and math, always look for the underlying pattern before you start calculating. The beauty of the solution is often hidden in the symmetry of the problem. The final answer is 11.