How to I approach this problem? Trigonometric or Algebraic computations?  The radicals as denominator is not all that encouraging too and I don't see any promising method to evaluate it. 

Solve for $n$ in the equation below:
  $$\frac{\sin\left(\frac{90^\circ}{2^n}\right)}{\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}_{n \text{ radicals}}}=\frac{-1+\sqrt2}{2}$$

 A: Use the half-angle formula:
$$\cos{x}=2\cos^2\frac{x}{2}-1$$
So
$$2+2\cos{x}=4\cos^2\frac{x}{2}$$
Then, for $0\leq x\leq\frac{\pi}{2}$,
$$2\cos\frac{x}{2}=\sqrt{2+2\cos x}$$
So
$$2\cos\frac{x}{2^{k+1}}=\sqrt{2+2\cos \frac{x}{2^k}}=\sqrt{2+\sqrt{2+2\cos \frac{x}{2^{k-1}}}}=\ldots$$
We have the denominator
$${\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}_{n \text{ radicals}}}$$
and we know that $\cos \pi =0$, so
$$2\cos\frac{\pi}{2^{n+1}}={\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}_{n \text{ radicals}}}$$
Hence, the LHS of the expression becomes
$$\frac{\sin\left(\frac{90^\circ}{2^n}\right)}{\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}_{n \text{ radicals}}}=\frac{\sin\left(\frac{\pi}{2^{n+1}}\right)}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}=\frac12\tan{\frac{\pi}{2^{n+1}}}$$
So we evaluate
$$\frac12\tan{\frac{\pi}{2^{n+1}}}=\frac{-1+\sqrt2}{2}$$
which simplifies to
$$\tan{\frac{\pi}{2^{n+1}}}=\sqrt2-1$$
$n$ can be easily found from here.
A: Using this or Proof of an equality involving cosine $\sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{2}}}}\ =\ 2\cos (\pi/2^{n+1})$,
$$\dfrac{\sin\dfrac\pi{2^{n+1}}}{2\cos\dfrac\pi{2^{n+1}}}=\dfrac{\sqrt2-1}2$$
$$\iff\tan\dfrac\pi{2^{n+1}}=\csc\dfrac\pi4-\cot\dfrac\pi4=\cdots=\tan\dfrac\pi8$$
