Problem source

Show that
\[
\cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)=\frac{\sin\alpha}{4\sin\left(\dfrac{\alpha}{4}\right)}\,,
\]
where $\alpha\neq k\pi$, $k$ is an integer. 

Prove that, for such $\alpha$, 
\[
\cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)=\frac{\sin\alpha}{2^{n}\sin\left(\dfrac{\alpha}{2^{n}}\right)}\,,
\]
where $n$ is a positive integer. 

Deduce that
\[
\alpha=\frac{\sin\alpha}{\cos\left(\dfrac{\alpha}{2}\right)\cos\left(\dfrac{\alpha}{4}\right)\cos\left(\dfrac{\alpha}{8}\right)\cdots}\,,
\]
and hence that
\[
\frac{\pi}{2}=\frac{1}{\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}\cdots}\,.
\]

Solution source

\begin{align*}
&&\sin \alpha &= 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \\
&&&= 4 \sin \frac{\alpha}{4} \cos \frac{\alpha}{4} \cos \frac{\alpha}{2} \\
\Rightarrow && \frac{\sin \alpha}{4 \sin \frac{\alpha}{4}} &= \cos \frac{\alpha}{2} \cos \frac{\alpha}{4}
\end{align*}

We proceed by induction on $n$. Clearly this is true for $n = 1$ (as we just established). Assume it is true for $n=k$. Then:

\begin{align*}
&& \frac{\sin \alpha}{2^n \sin \frac{\alpha}{2^n}} &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right) \\
\Rightarrow && \frac{\sin \alpha}{2^{n+1} \sin \frac{\alpha}{2^{n+1}} \cos \frac{\alpha}{2^{n+1}}}  &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right) \\
\Rightarrow && \frac{\sin \alpha}{2^{n+1} \sin \frac{\alpha}{2^{n+1}} }  &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)\cos \left ( \frac{\alpha}{2^{n+1}} \right) \\
\end{align*}
Therefore it is true for $n=k+1$ Therefore since it is true for $n=1$ and if it is true for $n=k$ it is also true for $n=k+1$ by the principle of mathematical induction it is true for all $n \geq 1$


\begin{align*}
\lim_{n \to \infty} \frac{\sin \alpha}{\cos\left(\frac{\alpha}{2}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)} &= \lim_{n \to \infty} 2^n \sin \frac{\alpha}{2^n} \\
&= \lim_{n \to \infty} \alpha \frac{\sin \frac{\alpha}{2^n}}{\frac{\alpha}{2^n}} \\
&= \alpha \lim_{t \to 0} \frac{\sin t}{t} \\
&= \alpha
\end{align*}

When $\alpha = \frac{\pi}{2}$ notice that $\sin \alpha =1$, $\cos \frac{\alpha}{2} = \sqrt{\frac12}$ and $2\cos^2 \frac{\alpha}{2^{n+1}}-1 = \cos \frac{\alpha}{2} \Rightarrow \cos \frac{\alpha}{2^{n+1}} = \sqrt{\frac12 + \cos \frac{\alpha}{2^n}}$ exactly the series we see.