Problem source

Let $a_{1}=\cos x$ with $0 < x < \pi/2$
and let $b_{1}=1$. Given that
\begin{eqnarray*}
a_{n+1}&=&{\textstyle \frac{1}{2}}(a_{n}+b_{n}),\\[2mm]
b_{n+1}&=&(a_{n+1}b_{n})^{1/2},
\end{eqnarray*}
find $a_{2}$ and $b_{2}$ and show that
\[a_{3}=\cos\frac{x}{2}\cos^{2}\frac{x}{4}
\ \quad\mbox{and}\quad
\ b_{3}=\cos\frac{x}{2}\cos\frac{x}{4}.\] 
Guess general expressions for $a_{n}$ and $b_{n}$ (for $n\ge2$)
as products of cosines
and verify that they satisfy the given equations.

Solution source

\begin{array}{c|c|c}
n & a_n & b_n \\ \hline
1 & \cos x & 1 \\ \hline
2 & \frac12(1 + \cos x) & \sqrt{a_2} \\
&=\frac12(1+2\cos^2 \frac{x}{2}-1)& \sqrt{a_2} \\
&= \cos^2 \frac{x}{2} & \cos \frac{x}{2} \\ \hline
3 & \frac12(\cos^2 \frac{x}{2}+\cos \frac{x}{2}) & \sqrt{a_3\cos \frac{x}{2}} \\
&= \cos \frac{x}{2} \cdot \frac12 (\cos \frac{x}{2}+1) & \sqrt{a_3\cos \frac{x}{2}} \\
&= \cos \frac{x}{2} \cos^2 \frac{x}{4} & \cos \frac{x}{2} \cos \frac{x}{4}
\end{array}

Claim: $\displaystyle a_n = \cos \frac{x}{2^{n-1}}\prod_{k=1}^{n-1} \cos \frac{x}{2^k}$, $\displaystyle b_n = \prod_{k=1}^{n-1} \cos \frac{x}{2^k}$

Claim: $a_{n+1} = \frac12(a_n + b_n)$

Proof: 
\begin{align*}
&&  \frac12(a_n + b_n) &= \frac12 \left ( \cos \frac{x}{2^{n-1}}\prod_{k=1}^{n-1} \cos \frac{x}{2^k} + \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \right) \\
&&&= \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \frac12\left (\cos \frac{x}{2^{n-1}} + 1 \right) \\
&&&= \left (  \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \right) \cos^{2} \frac{x}{2^n} \\
&&&= \cos \frac{x}{2^n} \prod_{k=1}^{n} \cos \frac{x}{2^k} \\
&= a_{n+1}
\end{align*}

Claim: $b_{n+1} = \sqrt{a_{n+1}b_n}$ 
Proof:
\begin{align*}
&&  \sqrt{a_{n+1}b_n} &= \sqrt{ \cos \frac{x}{2^n} \prod_{k=1}^{n} \cos \frac{x}{2^k} \cdot \prod_{k=1}^{n-1} \cos \frac{x}{2^k} }\\
&&&= \prod_{k=1}^{n-1} \cos \frac{x}{2^k}  \sqrt{\cos ^2\frac{x}{2^{n}}} \\
&&&= \prod_{k=1}^{n} \cos \frac{x}{2^k}  \\
&&&= b_{n+1}
\end{align*}