Prove by induction that $\cos\frac{90^\circ}{2^n}=\frac 12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...\sqrt{2}}}}}$

Question

I have a little problem. So I am ask to prove this identity by induction for $n\geq 1$

$$\cos\frac{90^\circ}{2^n}=\frac 12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...\sqrt{2}}}}}$$

which is relatively straigthforward if you just use regular weak induction and then for the inductive step, apply half angle trig identity and multiply top and bottom by 2...until you prove if $n=k$ true then $n=k+1$ true. simple

But the second part of the question asks to use this identity to find

a) $\sin\frac{90^\circ}{2^n}$

b) hence, an expression converging to $\pi$

For a) I can only think of using the pythagorean identity $\sin^2 + \cos^2 = 1$ and then rearraging to make $\sin$ the subject where $\theta = \sqrt{1-\cos^2{\theta}}$ and $\theta$ = that monster up above. But wouldn't this require having to take the square of that monster? I do not know how to evaluate that monster up there for n terms.

as for part b) to find a limit to some number I can only think of somehow using that trig limit $\lim_{\theta\to 0}\frac{\sin\theta}{\theta} =1$. I also am aware that as $n$ is really big, the angle gets really small. So could I please have help? I have been stuck on this thingy for days. Thank you so much!

Answer 1

You can use Identity $\sin(2a)=2\sin(a)\cos(a)$
first step $n=k \to \sin\frac{90^\circ}{2^{k-1}}=2\sin\frac{90^\circ}{2^k}\cos\frac{90^\circ}{2^k}$ it gives you a recursive realation $$\sin\frac{90^\circ}{2^{k-1}}=2\sin\frac{90^\circ}{2^k}\times \frac 12\sqrt{\underbrace{2+\sqrt{2+\sqrt{2+\sqrt{...\sqrt{2}}}}}_{k-times}}\\ $$ so take $\sin\frac{90^\circ}{2^{k-1}}=f(k-1)$ and you will have $$f(k-1)=2\times \frac 12\sqrt{\underbrace{2+\sqrt{2+\sqrt{2+\sqrt{...\sqrt{2}}}}}_{k-times}}\times f(k)\\ $$so $$f(k-1)=\sqrt{\underbrace{2+\sqrt{2+\sqrt{2+\sqrt{...\sqrt{2}}}}}_{k-times}}\times f(k)\to \\ f(k)=\frac{1}{\sqrt{\underbrace{2+\sqrt{2+\sqrt{2+\sqrt{...\sqrt{2}}}}}_{k-times}}}f(k-1)$$ then it's enough to rationalize denominator / or use induction again

Answer 2

You could probably benefit from some compact notation, perhaps just defining the sequence

$a_0=\cos{\frac{\pi}{2}}=0$, and for $n\geq 1$:

$a_n=(1/2)\sqrt{2+2a_{n-1}}=\frac{\sqrt{2}}{2}\sqrt{1+a_{n-1}} =\cos{\frac{\pi}{2^{n+1}}}$

$b_n=\sqrt{1-a_n^2}=\sqrt{1-\frac{1+a_{n-1}}{2}}=\frac{\sqrt{2}}{2}\sqrt{1-a_{n-1}}=\sin{\frac{\pi}{2^{n+1}}}$

And $\frac{b_n}{\frac{\pi}{2^{n+1}}}\to 1$ means $2^{n+1}b_n\to \pi$, since as $n\to\infty$, $\frac{\pi}{2^{n+1}}\to 0$. The limit looks odd, like it can't possibly converge, but if the expression for $b_n$ is right then $b_n$ is small enough that it balances the $2^{n+1}$. Notice that the $a_n$ and $b_n$ expressions are just the half-angle formulas.

Khosrotash's answer gives the radical expression for $a_n$ neatly enough I think, which you can then plug neatly into the expression for $b_n$.