algebra precalculus - 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:

$$\frac{\sin\left(\frac{90^\circ}{2^n}\right)}{\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}_{n \text{ radicals}}}=\frac{-1+\sqrt2}{2}$$

Answer

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.