Animated Solution for Mathematics - Trigonometry: The expression $3[{\sin }^4(\frac{3\pi }{2}-\alpha )+{\sin }^4(3\pi +\alpha )]-2[{\sin }^6(\frac{\pi }{2}+\alpha )+{\sin }^6(5\pi -\alpha )]$ is equal to

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that looks like a fortress of complexity but is, in reality, a beautifully choreographed dance of identities.

When you first look at an expression like $3[{\sin }^4(\frac{3\pi }{2}-\alpha )+{\sin }^4(3\pi +\alpha )]-2[{\sin }^6(\frac{\pi }{2}+\alpha )+{\sin }^6(5\pi -\alpha )]$, it is natural to feel a moment of hesitation. But remember, in JEE Advanced, the most intimidating expressions are often the ones that hide the most elegant symmetries.

Taming the Quadrants

Our first mission is to simplify the 'allied angles'. Imagine yourself standing at the origin of the unit circle. We use the ASTC rule—All-Silver-Tea-Cups—to determine the sign of our functions.

Consider $\sin (\frac{3\pi }{2}-\alpha )$. We are in the third quadrant, where sine is negative. Because we are dealing with an odd multiple of $\frac{\pi }{2}$, the sine function transforms into cosine.

Thus, $\sin (\frac{3\pi }{2}-\alpha )=-\cos \alpha$. When we raise this to the fourth power, the negative sign disappears, leaving us with ${\cos }^4\alpha$.

Next, look at $\sin (3\pi +\alpha )$. This angle is coterminal with $\pi +\alpha$, landing us firmly in the third quadrant. Since $3\pi$ is an integer multiple of $\pi$, the function remains sine.

So, $\sin (3\pi +\alpha )=-\sin \alpha$. Again, the fourth power turns this into ${\sin }^4\alpha$.

The Power of Six

Now, let us tackle the second bracket. For $\sin (\frac{\pi }{2}+\alpha )$, we are in the second quadrant where sine is positive. The odd multiple of $\frac{\pi }{2}$ flips it to $\cos \alpha$.

Raising this to the sixth power gives us ${\cos }^6\alpha$. Finally, $\sin (5\pi -\alpha )$ is coterminal with $\pi -\alpha$, which is in the second quadrant. Sine remains sine, so we get ${\sin }^6\alpha$.

Our expression $E$ has now transformed into:

The Algebraic Alchemy

This is where the magic happens. We know that ${\sin }^2\alpha +{\cos }^2\alpha =1$. We can use this to rewrite our powers.

For the fourth powers, we use the identity $a^2+b^2=(a+b{)}^2-2ab$. By setting $a={\sin }^2\alpha$ and $b={\cos }^2\alpha$, we get:

For the sixth powers, we use the cubic identity $a^3+b^3=(a+b{)}^3-3ab(a+b)$:

Now, we substitute these back into our expression $E$:

Watch closely as we distribute the constants:

The terms involving ${\sin }^2\alpha {\cos }^2\alpha$ cancel out with perfect precision! We are left with $3-2=1$.

Isn't it beautiful? No matter what value of $\alpha$ you choose, the expression remains constant. You have successfully navigated the complexity and arrived at the truth.