Animated Solution for Mathematics - Trigonometry: $3(\sin x-\cos x{)}^4+6(\sin x+\cos x{)}^2+4({\sin }^{6}x+{\cos }^{6}x)=$

Welcome, future IITian. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of powers and trigonometric functions.

The expression is $3(\sin x-\cos x{)}^4+6(\sin x+\cos x{)}^2+4({\sin }^{6}x+{\cos }^{6}x)$.

When you first see this, your instinct might be to panic. You might think, 'Do I really need to expand a power of four and a power of six?' The answer is a resounding no.

In the world of JEE Advanced, we do not brute-force our way through problems; we use elegance, symmetry, and the 'divide and conquer' strategy.

The Divide and Conquer Strategy

Imagine you are standing before a massive fortress. You don't attack the whole wall at once; you find the weak points.

Here, our weak points are the three distinct components of the expression. We will isolate them, simplify them individually, and then bring them back together.

Our goal is to express everything in terms of a single, simple trigonometric function: $\sin 2x$.

Unlocking the Squares

Let's look at the first two terms: $(\sin x-\cos x{)}^4$ and $(\sin x+\cos x{)}^2$. Notice the base: $(\sin x\pm \cos x)$.

If we square these, something magical happens. Recall the identity $(a\pm b{)}^2=a^2+b^2\pm 2ab$.

Applying this to $(\sin x-\cos x{)}^2$, we get ${\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x$. Since ${\sin }^{2}x+{\cos }^{2}x=1$ and $2\sin x\cos x=\sin 2x$, this simplifies to $1-\sin 2x$.

Similarly, $(\sin x+\cos x{)}^2$ becomes $1+\sin 2x$. Now, our first term becomes $3(1-\sin 2x{)}^2$ and our second term becomes $6(1+\sin 2x)$.

The Power of Six Trap

Now, consider the third term: $4({\sin }^{6}x+{\cos }^{6}x)$. Powers of 6 are just cubes of squares, so we write this as $({\sin }^{2}x{)}^3+({\cos }^{2}x{)}^3$.

Using the sum of cubes identity $a^3+b^3=(a+b)(a^2-ab+b^2)$, where $a={\sin }^{2}x$ and $b={\cos }^{2}x$, we get:

Since ${\sin }^{2}x+{\cos }^{2}x=1$, we are left with ${\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x$. By adding and subtracting $2{\sin }^{2}x{\cos }^{2}x$, we rewrite ${\sin }^{4}x+{\cos }^{4}x$ as:

Putting it all together, the term simplifies to $1-3{\sin }^{2}x{\cos }^{2}x$. Since ${\sin }^{2}x{\cos }^{2}x=\frac{1}{4}{\sin }^{2}2x$, this term becomes:

The Grand Unification

Now, let's assemble our pieces:

Expanding the first term:

Adding the second and third terms:

Look at the cancellation! The $-6\sin 2x$ and $+6\sin 2x$ vanish. The $+3{\sin }^{2}2x$ and $-3{\sin }^{2}2x$ vanish.

We are left with $3+6+4=13$. The entire expression is independent of $x$, resulting in the constant value 13.