Use a trig substitution (Half-Angle) in $a_n=2^\frac{n+1}{2}\sqrt{2^n-\sqrt{4^n-a_{n-1}^2}} \ \ \ \forall n\gt1$

Question

I'm given $$a_1=2\sqrt{2}$$$$a_n=2^\frac{n+1}{2}\sqrt{2^n-\sqrt{4^n-a_{n-1}^2}} \ \ \ \forall n\gt1$$

I've tried finding $a_1,a_2,a_3,....$ to try and find a pattern, but it gives no simple pattern to follow. I've tried using generating fuctions, but I get $a_n^2$ which doesn't turn out very nice.

A friend told me that using a trig identity works nicely. He even told me it was the half-angle identity, but I have no Idea how the half angle identity fits in.

I know the Half-Angle Identities are $$\sin(\frac x2)=+-\sqrt{\frac{1-cos(x)}2}$$$$\cos(\frac x2)=+-\sqrt{\frac{1+cos(x)}2}$$ I would assume it is the $\sin(\frac x2)$ since we are dealing with subtraction, but I don't know how to start this problem.

If someone could get me started it'd be much appreciated! I don't want the answer just some help.

Answer 1

Hint: How about $a_n = 2^{n+1}\sin \theta_n$