Animated Solution for Mathematics - Trigonometry: If $f(x)={\sin }^{2}x+{\sin }^2(x+\pi /3)+\cos x\cos (x+\pi /3)$ and $g(5/4)=1$, then $(g\circ f)(x)=.........$

Welcome, future engineers. Today, we are going to solve a problem that looks intimidating at first glance but is actually a masterclass in elegance.

We are presented with a function $f(x)={\sin }^{2}x+{\sin }^2(x+\frac{\pi }{3})+\cos x\cos (x+\frac{\pi }{3})$.

At first, you might feel the urge to panic. It looks like a mess of trigonometric terms. But stop. Take a breath.

Look at the second part of the question: $g(5/4)=1$. This is not just information; it is a massive clue. It is the examiner whispering to you, "The function $f(x)$ is not a variable; it is a constant, and that constant is $5/4$." Our job is to prove it.

Breaking Down the Barriers

To simplify $f(x)$, we need to dismantle the terms containing $(x+\frac{\pi }{3})$. We cannot work with these compound angles directly.

We must use our compound angle identities. Recall that $\sin (A+B)=\sin A\cos B+\cos A\sin B$ and $\cos (A+B)=\cos A\cos B-\sin A\sin B$.

Here, our $A$ is $x$ and our $B$ is $\frac{\pi }{3}$. We know that $\sin (\frac{\pi }{3})=\frac{\sqrt{3}}{2}$ and $\cos (\frac{\pi }{3})=\frac{1}{2}$.

Let us substitute these values into our expression. The function becomes:

The Expansion and the Magic Cancellation

Now, we expand the squared term using the identity $(a+b{)}^2=a^2+b^2+2ab$. This gives us:

Next, we distribute the $\cos x$ in the third term:

Look closely at these two results. Do you see it? The term $+\frac{\sqrt{3}}{2}\sin x\cos x$ from the expansion and the term $-\frac{\sqrt{3}}{2}\sin x\cos x$ from the distribution are identical in magnitude but opposite in sign.

They cancel out perfectly! This is the "magic" of JEE problems—the complexity vanishes when you apply the right tools.

The Final Identity

With the cross terms gone, we are left with:

Let us group the ${\sin }^{2}x$ terms and the ${\cos }^{2}x$ terms. For ${\sin }^{2}x$, we have $(1+\frac{1}{4})=\frac{5}{4}$. For ${\cos }^{2}x$, we have $(\frac{3}{4}+\frac{1}{2})=\frac{5}{4}$.

Thus, $f(x)=\frac{5}{4}{\sin }^{2}x+\frac{5}{4}{\cos }^{2}x$. Factoring out the $\frac{5}{4}$, we get:

Since ${\sin }^{2}x+{\cos }^{2}x=1$, our function simplifies to $f(x)=\frac{5}{4}$.

The Beauty of Simplicity

We have proven that $f(x)$ is always $5/4$. Now, the composite function $(g\circ f)(x)$ is simply $g(f(x))$, which is $g(5/4)$.

Since the problem explicitly told us $g(5/4)=1$, our final answer is 1.

This problem teaches us that even when a function looks terrifying, a systematic application of fundamental identities will always lead to the truth. Keep practicing, keep simplifying, and keep falling in love with the process.