Animated Solution for Mathematics - Trigonometry: Let $f(\theta )=3\left({\sin }^4\left(\frac{3\pi }{2}-\theta \right)+{\sin }^4(3\pi +\theta )\right)-2(1-{\sin }^{2}2\theta )$ and $S=\{\theta \in [0,\pi ]:f^{\prime }(\theta )=-\frac{\sqrt{3}}{2}\}$. If $4\beta ={\sum }_{\theta \in S}\theta$, then $f(\beta )$ is equal to

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to dissect a problem that, at first glance, looks like a tangled mess of trigonometric powers.

You see an expression like $f(\theta )=3\left({\sin }^4\left(\frac{3\pi }{2}-\theta \right)+{\sin }^4(3\pi +\theta )\right)-2(1-{\sin }^{2}2\theta )$ and your instinct might be to panic. But I want you to take a deep breath. In mathematics, as in life, complexity is often just a mask for a simpler, more elegant truth waiting to be revealed.

Our first mission is to simplify the function. We start by looking at the arguments inside the sine functions.

Recall your reduction formulas: $\sin (\frac{3\pi }{2}-\theta )=-\cos \theta$ and $\sin (3\pi +\theta )=-\sin \theta$. Because these terms are raised to the fourth power, the negative signs vanish into thin air!

We are left with:

Now, look at the term ${\cos }^4\theta +{\sin }^4\theta$. This is a classic identity. We know that $({\sin }^2\theta +{\cos }^2\theta {)}^2={\sin }^4\theta +{\cos }^4\theta +2{\sin }^2\theta {\cos }^2\theta$.

Since ${\sin }^2\theta +{\cos }^2\theta =1$, we can write ${\sin }^4\theta +{\cos }^4\theta =1-2{\sin }^2\theta {\cos }^2\theta$. Using the double angle identity $\sin 2\theta =2\sin \theta \cos \theta$, we see that $2{\sin }^2\theta {\cos }^2\theta =\frac{1}{2}{\sin }^{2}2\theta$.

Thus, the expression becomes $1-\frac{1}{2}{\sin }^{2}2\theta$.

Substituting this back into our function, we get:

Since $1-{\sin }^{2}2\theta ={\cos }^{2}2\theta$, we can unify the terms. After a bit of algebraic housekeeping, we arrive at the beautiful, compact form:

Isn't that satisfying? We have transformed a daunting expression into a simple cosine wave. Now, finding the derivative $f^{\prime }(\theta )$ is a breeze. The constant $\frac{5}{4}$ vanishes, and the derivative of $-\frac{1}{4}\cos 4\theta$ becomes $\sin 4\theta$.

We are given $f^{\prime }(\theta )=-\frac{\sqrt{3}}{2}$, which means $\sin 4\theta =-\frac{\sqrt{3}}{2}$. Since $\theta \in [0,\pi ]$, our argument $4\theta$ lives in the interval $[0,4\pi ]$.

We are looking for the points where the sine function hits $-\frac{\sqrt{3}}{2}$ across two full cycles. These occur at $4\theta =\frac{4\pi }{3},\frac{5\pi }{3},\frac{10\pi }{3},$ and $\frac{11\pi }{3}$.

Summing these values gives us $4\beta =\frac{4\pi +5\pi +10\pi +11\pi }{3}=\frac{30\pi }{3}=10\pi$. However, calculating the average value $\beta$ for the sum of roots, we find $4\beta =\frac{5\pi }{2}$.

Finally, we evaluate $f(\beta )=\frac{5}{4}-\frac{1}{4}\cos 4\beta$. Substituting $4\beta =\frac{5\pi }{2}$, we get $\cos (\frac{5\pi }{2})=0$.

Therefore, the final result is:

We have navigated the storm of trigonometry and emerged on the other side with a clean, precise answer. Remember, the key to JEE Advanced is not just knowing the formulas, but having the patience to simplify the world around you until the answer reveals itself.