Animated Solution for Mathematics - Complex Numbers: The value of ${\left(\frac{1+\sin \frac{2\pi }{9}+i\cos \frac{2\pi }{9}}{1+\sin \frac{2\pi }{9}-i\cos \frac{2\pi }{9}}\right)}^3$ is:

Welcome, future engineers! Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of trigonometry and complex numbers.

You see an expression like:

Your instinct might be to panic. But I want you to take a deep breath. In JEE Advanced, the most intimidating problems are often the ones that hide the simplest symmetries. We are not going to brute-force this; we are going to dance through it.

The Transformation

Our first step is to simplify the base of this expression. Let $\theta =\frac{2\pi }{9}$. The expression is:

Notice something? The numerator and denominator are complex conjugates. This is a massive hint. However, the standard Euler form $e^{iA}=\cos A+i\sin A$ requires the real part to be cosine and the imaginary part to be sine. We have the opposite!

This is where we use our complementary angle identities. We know that $\sin \theta =\cos (\frac{\pi }{2}-\theta )$ and $\cos \theta =\sin (\frac{\pi }{2}-\theta )$. Let $A=\frac{\pi }{2}-\theta$.

Suddenly, our fraction transforms into:

Now, it looks like a problem we can actually solve.

The Elegance of Half-Angles

Now, we bring in the heavy artillery: the half-angle identities. We know that $1+\cos A=2{\cos }^2\frac{A}{2}$ and $\sin A=2\sin \frac{A}{2}\cos \frac{A}{2}$.

When we substitute these into our fraction, the magic happens:

Look at that! The $2\cos \frac{A}{2}$ terms cancel out completely. We are left with:

Using Euler's formula, this is simply:

We have reduced a terrifying fraction to a single exponential term. This is the power of mathematical foresight.

The Final Rotation

We defined $A=\frac{\pi }{2}-\theta$. With $\theta =\frac{2\pi }{9}$, we calculate:

So, our base is $e^{i\frac{5\pi }{18}}$. Now, we apply the cube power:

We are almost there. We expand $e^{i\frac{5\pi }{6}}$ into $\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}$. Since $\frac{5\pi }{6}$ is in the second quadrant:

This gives us the final result:

Even if the options have a slight typo, the structure is clear. You have mastered the logic. Trust your math, and you will always find the way.