MHB Finding an Alternative Solution to Doubtful Numerator Steps

[DaalChawal]

Doubt in marked step(Numerator) that how it came from first?
Is there any alternative solution for this?

 

[I like Serena]

We generally have $\sin^2x + \cos^2 x=1$.
So $\sin^2x - i^2 \cos^2 x=1$ as well.
Now substitute $x=\frac{2\pi}{9}$.

 

[DaalChawal]

Thanks